Gravitation - Class 11 Physics - Chapter 7 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Gravitation | NCERT | Physics | Class 11
The first man to land on the Moon was
A. Edwin Aldrin
B. Buzz Aldrin
C. Neil Armstrong
D. Michael Collins
The correct answer is C. Neil Armstrong.
Neil Armstrong was the first man to step foot on the Moon.
A hypothetical planet has density $r$, radius $R$, and surface gravitational acceleration $g$. If the radius of the planet were doubled, but the planetary density stayed the same, the acceleration due to gravity at the planet's surface would be:
A $4 \mathrm{~g}$
B $2 \mathrm{~g}$
C $\mathrm{g}$
D $\frac{\mathrm{g}}{2}$.
The correct option is B $2g$.
For the first case: The acceleration due to gravity $g$ on the surface of the planet can be expressed by: $$ g = \frac{G \cdot M}{R^2} $$ where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet. The mass of the planet, with density $\rho$ and volume $\frac{4}{3}\pi R^3$, is: $$ M = \rho \cdot \frac{4}{3} \pi R^3 $$ Substituting that into the formula for $g$ gives: $$ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} = \frac{4}{3} \pi G \rho R $$
For the second case: If the radius, $R$, is doubled to $2R$ while keeping the density, $\rho$, constant:
The new mass, $M'$, of the planet would be: $$ M' = \rho \cdot \frac{4}{3} \pi (2R)^3 = 8 \rho \cdot \frac{4}{3} \pi R^3 $$
The new gravitational acceleration, $g'$, on the surface would be: $$ g' = \frac{G \cdot M'}{(2R)^2} = \frac{G \cdot 8 \rho \cdot \frac{4}{3} \pi R^3}{4R^2} = 2 \left(\frac{4}{3} \pi G \rho R\right) $$ indicating: $$ g' = 2g $$
Therefore, the new surface gravitational acceleration is twice the original value, which is option B $2g$.
What is the effect of altitude on the value of $g$?
Gravity ($g$) is inversely proportional to the square of the distance from the center of the Earth. As altitude increases, this distance increases, leading to a reduction in the gravitational force. Mathematically, this is represented as:
$$ g' = g\left(1 + \frac{h}{R}\right)^{-2} $$
where $g'$ is the gravitational acceleration at altitude $h$, $R$ is the Earth's radius, and $g$ is the gravitational acceleration at sea level.
For small values of $h$ compared to $R$, we can simplify the equation using the binomial expansion, which gives:
$$ g' = g\left(1 - \frac{2h}{R}\right) $$
Therefore, it is evident that as altitude increases, the value of $g$ decreases, thereby making gravity weaker.
Who among the following was not an explorer?
A. Hrafna-Flóki Vilgerðarson
B. Isaac Newton
C. Christopher Columbus
D. Ferdinand Magellan
The correct answer to the question is B. Isaac Newton.
Isaac Newton was primarily known for his ground-breaking contributions as a mathematician, astronomer, and physicist and is considered one of the most influential scientists in history. Unlike Hrafna-Flóki Vilgerðarson, Christopher Columbus, and Ferdinand Magellan, who were noted explorers, Newton did not engage in geographical explorations. Notably, Hrafna-Flóki Vilgerðarson is recognized for discovering Iceland, while Columbus and Magellan embarked on voyages that significantly contributed to the European exploration of the world.
Choose the correct alternatives.
(i) Acceleration due to gravity increases/decreases with increasing altitude.
(ii) Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
(iii) Acceleration due to gravity is independent of the mass of the Earth/mass of the body.
(iv) The formula $-G M m\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right)$ is more/less accurate than the formula $\mathrm{mg}\left(r_{2}-r_{1}\right)$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the Earth.
(i) Acceleration due to gravity increases/decreases with increasing altitude.
When we consider the Earth with radius $R$:
At the surface, the acceleration due to gravity is $g$.
At an altitude $h$ above the Earth's surface, the acceleration due to gravity is $g'$.
The relationship between $g$ and $g'$ at height $h$ is given by:
$$ g' = g \left( 1 - \frac{2h}{R} \right) $$
As the altitude $h$ increases, the term $\frac{2h}{R}$ also increases, making $g'$ decrease. Therefore, acceleration due to gravity decreases with increasing altitude.
(ii) Acceleration due to gravity increases/decreases with increasing depth.
At the Earth's surface, acceleration due to gravity is $g$, and at a depth $d$ below the surface, the acceleration due to gravity is $g'$.
Using the relation for gravity inside the Earth (assuming uniform density):
$$ g' = g \left( 1 - \frac{d}{R} \right) $$
As the depth $d$ increases, the term $\frac{d}{R}$ increases, making $g'$ decrease. Thus, acceleration due to gravity decreases with increasing depth.
(iii) Acceleration due to gravity is independent of the mass of the Earth/mass of the body.
Acceleration due to gravity, $g$, is given by:
$$ g = \frac{GM}{R^2} $$
Where:
$G$ is the gravitational constant,
$M$ is the mass of the Earth,
$R$ is the radius of the Earth.
The formula shows $g$ depends on the mass of the Earth and the radius of the Earth, but it does not depend on the mass of the body. Therefore, acceleration due to gravity is independent of the mass of the body.
(iv) The formula $-GMm \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$ is more/less accurate than the formula $mg(r_2 - r_1)$ for the difference in potential energy between two points $r_2$ and $r_1$ distance away from the centre of the Earth.
For evaluating potential energy differences, we use:
The exact formula: $$ U_b - U_a = -GMm \left( \frac{1}{r_2} - \frac{1}{r_1} \right) $$
An approximation for small height differences near the Earth's surface: $$ \Delta U = mg(r_2 - r_1) $$
The exact formula considers the variability of gravity with distance, making it more accurate than the approximate linear formula. Hence, the formula $-GMm \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$ is more accurate.
In summary, the correct alternatives are:
Decreases with increasing altitude.
Decreases with increasing depth.
Independent of the mass of the body.
More accurate.
The masses of two planets are in the ratio 1: 2. Their radii are in the ratio 1: 2. The acceleration due to gravity on the planets is in the ratio.
1: 2
2: 1
3: 5
5: 3
The correct answer is 2:1.
Given the formula for the acceleration due to gravity: $$ g = \frac{GM}{R^2} $$
For the first planet: $$ g_1 = \frac{GM_1}{R_1^2} $$
For the second planet: $$ g_2 = \frac{GM_2}{R_2^2} $$
It is provided that the masses and radii of the two planets follow these ratios: $$ M_2 = 2M_1 $$ $$ R_2 = 2R_1 $$
By substituting these values into the gravity formulas: $$ g_1 = \frac{GM_1}{R_1^2} $$ $$ g_2 = \frac{G(2M_1)}{(2R_1)^2} = \frac{2GM_1}{4R_1^2} = \frac{GM_1}{2R_1^2} $$
Thus, $$ \frac{g_1}{g_2} = \frac{\frac{GM_1}{R_1^2}}{\frac{GM_1}{2R_1^2}} = 2:1 $$
Therefore, the ratio of the acceleration due to gravity on the two planets is 2:1.
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Ask Chatterbot AINCERT Solutions - Gravitation | NCERT | Physics | Class 11
Answer the following :
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun's pull is greater than the moon's pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon's pull is greater than the tidal effect of sun. Why ?
(a) Shielding from Gravitational Influence
Gravitational forces cannot be shielded. Placing a body inside a hollow sphere does not shield it from the gravitational influence of nearby matter. According to the principles discussed,
$$ \text{The gravitational force on a particle inside a uniform spherical shell is zero. However, the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside. Gravitational shielding is not possible.} $$
(b) Detecting Gravity inside a Large Space Station
In a small spacecraft orbiting Earth, the astronaut experiences weightlessness because both the astronaut and the spacecraft are in free-fall towards Earth. This is true regardless of the spacecraft's size. Inside a larger space station, the same principle applies, but small variations due to the station's size might allow the astronaut to detect slight differences in gravitational force if the structure is sufficiently large (i.e., if there are significant distances within the space station).
(c) Greater Tidal Effect of the Moon than the Sun
Although the gravitational pull of the Sun on Earth is greater than that of the Moon, the tidal effect depends on the difference in gravitational pull across the Earth's diameter. The tidal force is proportional to the gradient (change) of the gravitational field, not just its strength. The Moon, being much closer to the Earth than the Sun, creates a stronger differential force across the Earth’s diameter, leading to a greater tidal effect.
In mathematical terms, the tidal force ( F_t ) is given by:
$$ F_t \propto \frac{2GMm}{r^3} $$
where ( G ) is the gravitational constant, ( M ) is the mass causing the tidal effect, ( m ) is the mass being affected, and ( r ) is the distance to the mass causing the tidal effect. For the Moon (( r ) is small) compared to the Sun (( r ) is large), the tidal effect ( \propto \frac{1}{r^3} ) is greater for the Moon.
Choose the correct alternative :
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula $-G M m\left(1 / r_{2}-1 / r_{1}\right)$ is more/less accurate than the formula $m g\left(r_{2}-r_{1}\right)$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the earth.
(a) Acceleration due to gravity increases/decreases with increasing altitude.
From the chapter, the formula for gravitational acceleration at height $h$ above the Earth's surface is: [ g(h) = \frac{G M_E}{(R_E + h)^2} ] Since ((R_E + h)) increases as (h) increases, (g(h)) decreases. So, acceleration due to gravity decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
From the chapter, the formula for gravitational acceleration at a depth (d) below the Earth's surface is: [ g(d) = g \left(1 - \frac{d}{R_E}\right) ] As (d) increases, the term (\left(1 - \frac{d}{R_E}\right)) decreases. So, acceleration due to gravity decreases with increasing depth.
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
From the chapter, the formula for gravitational acceleration is: [ g = \frac{G M_E}{R_E^2} ] This formula depends on (M_E) (mass of the Earth) and (R_E) (radius of the Earth). However, it is independent of the mass of the body experiencing the acceleration. So, acceleration due to gravity is independent of the mass of the body.
(d) The formula ( -G M m \left(1 / r_2 - 1 / r_1 \right) ) is more/less accurate than the formula ( m g \left(r_2 - r_1 \right) ) for the difference of potential energy between two points ( r_2 ) and ( r_1 ) distance away from the centre of the earth.
[ -G M m \left( \frac{1}{r_2} - \frac{1}{r_1} \right) ] This formula takes into account the variation of the gravitational force with distance from the Earth's centre and is valid for all distances.
[ m g \left( r_2 - r_1 \right) ] This formula assumes the gravitational acceleration (g) to be constant, which is an approximation valid only for small distances ( (r_2 - r_1) ) near the Earth's surface.
So, the formula (-G M m \left(1 / r_2 - 1 / r_1 \right) ) is more accurate.
Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?
The value of ( \left(\frac{1}{4}\right)^{\frac{1}{3}} ) is approximately 0.63.
Thus, the semi-major axis ( R_{P} ) of the new planet's orbit is: $$ R_{P} \approx 0.63 \times R_{E} $$
Therefore, the orbital size of the new planet is approximately 63% of the Earth's orbital size.
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is $4.22 \times 10^{8} \mathrm{~m}$. Show that the mass of Jupiter is about one-thousandth that of the sun.
The ratio ( \frac{M_J}{M_{\odot}} ) is approximately ( 9.573 \times 10^{-4} ).
This value is very close to ( 1 \times 10^{-3} ) or one-thousandth. Therefore, the calculated mass of Jupiter is about one-thousandth that of the Sun, just as stated.
Let us assume that our galaxy consists of $2.5 \times 10^{11}$ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be $10^{5} \mathrm{ly}$.
Calculation Results
Orbital Velocity ( v ): $$ v = 264,761 \text{ m/s} $$
Orbital Period ( T ): $$ T = 1.123 \times 10^{16} \text{ seconds} $$ $$ T \approx 3.557 \times 10^{8} \text{ years} $$
Explanation
The orbital period ( T ) of the star at a distance of 50,000 light-years from the galactic center is approximately 355.7 million years.
This is the time it will take for the star to complete one full revolution around the center of the galaxy.
Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth's gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth's influence.
Part (a)
The total energy ( E ) of an orbiting satellite is given by the sum of its kinetic energy ( K.E. ) and its potential energy ( P.E. ). For an orbiting satellite:
[ \text{Total Energy},, E = K.E. + P.E. ]
From the relevant section of the chapter, we know: [ K.E. = \frac{G M_E m}{2(R_E + h)} ] [ P.E. = -\frac{G M_E m}{R_E + h} ]
So, [ E = \frac{G M_E m}{2(R_E + h)} - \frac{G M_E m}{R_E + h} = -\frac{G M_E m}{2(R_E + h)} ]
This shows: [ E = \text{negative of half its potential energy} ]
Thus, if the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic energy.
Answer to part (a): kinetic energy
Part (b)
For a satellite in orbit, it already has some kinetic energy due to its motion. The total energy to remove it from Earth's influence is the work needed to counteract its current total energy (which includes both kinetic and potential energy).
However, for a stationary object at the same height, it has no initial kinetic energy, only potential energy.
Therefore:
Energy required to remove a satellite already in motion: Work against its negative total energy ( E )
Energy required to project a stationary object from the same height: Work against its potential energy ( P.E. )
Since the satellite has kinetic energy contributing to its total negative energy, more energy is required to remove the stationary object as the satellite in orbit already has kinetic energy working in its favour.
Answer to part (b): more
Final Answer:
(a) kinetic
(b) more
Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?
The escape speed of a body from the Earth can be analyzed using the escape speed formula:
$$ v_e = \sqrt{\frac{2GM_E}{R_E}} $$
where:
( v_e ) is the escape speed,
( G ) is the universal gravitational constant,
( M_E ) is the mass of the Earth,
( R_E ) is the radius of the Earth.
Let's analyze each factor:
Mass of the body: The formula does not include the mass of the body; hence, \textbf{the escape speed does not depend on the mass of the body}.
Location from where it is projected: The escape speed formula remains the same for any location on the surface of the Earth assuming ( R_E ) is the radius from the Earth's center to the surface. If the location is not on the surface, we would use the formula considering ( r ) as ( R_E + h ), where ( h ) is the height from the surface. Hence, \textbf{the escape speed does depend on the location in terms of height above the Earth's surface}.
Direction of projection: The escape speed formula derives from the requirement to overcome the gravitational potential energy, and it is independent of the direction of projection. \textbf{The escape speed does not depend on the direction of projection}.
Height of the location from where the body is launched: If the body is launched from a height ( h ) above the Earth's surface, the escape speed is modified and given by:
$$ v_e = \sqrt{\frac{2GM_E}{R_E + h}} $$
Thus, \textbf{the escape speed does depend on the height of the location from where the body is launched}.
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Comet Orbit Characteristics
A comet in a highly elliptical orbit around the sun presents some interesting characteristics. Here's how each of the quantities compare throughout its orbit:
(a) Linear Speed
The comet does not have a constant linear speed. According to Kepler's second law (the Law of Areas), a line joining a planet and the Sun sweeps out equal areas during equal intervals of time. Therefore, the comet moves faster when it is nearer to the Sun (perihelion) and slower when it is farther from the Sun (aphelion).
(b) Angular Speed
The comet does not have a constant angular speed. This is also due to Kepler's second law. Angular speed varies as the comet gets closer or further away from the Sun.
(c) Angular Momentum
The comet does have a constant angular momentum. This is a consequence of the conservation of angular momentum. In a gravitational orbit (central force), angular momentum remains constant.
(d) Kinetic Energy
The comet does not have constant kinetic energy. The kinetic energy changes because the linear speed of the comet changes as it moves closer to or further from the Sun. At perihelion, the kinetic energy is higher, and at aphelion, it is lower.
(e) Potential Energy
The comet does not have constant potential energy. The potential energy varies with the distance from the Sun: it is more negative when the comet is closer to the Sun (due to the stronger gravitational pull) and less negative when it is farther away.
(f) Total Energy
The comet does have a constant total energy throughout its orbit, provided we neglect any mass loss. The total energy (kinetic energy + potential energy) in an elliptical orbit remains constant, even though both the kinetic and potential energies separately vary with the position of the comet.
Summary
Linear Speed: Not constant.
Angular Speed: Not constant.
Angular Momentum: Constant.
Kinetic Energy: Not constant.
Potential Energy: Not constant.
Total Energy: Constant.
Hence, angular momentum and total energy are constant throughout the comet's elliptical orbit, while others vary.
Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.
In space, astronauts commonly experience several symptoms due to the effects of microgravity. Among the options given:
Swollen face (b): In microgravity, fluids in the body are redistributed and often move towards the upper body and head, causing facial puffiness or swelling.
Headache (c): This can occur due to fluid shift and other factors associated with the space environment.
Orientational problem (d): In the absence of gravity, the senses that help with balance and orientation can become confused, leading to orientation problems.
Swollen feet (a) is less likely because, in microgravity, fluids tend to move away from the lower extremities.
Thus, all of the following symptoms – swollen face, headache, and orientational problems – are likely to afflict an astronaut in space.
In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0 .
At the center of a hemispherical shell, the gravitational force will be directed towards the shell, as the mass is uniformly distributed over the surface of the hemisphere. This results in a net gravitational force pointing in the direction normal to the flat face of the hemisphere.
To find the correct answer for the direction of the gravitational intensity:
(i) a: Diagonally up and to the right.
(ii) b: Vertically up.
(iii) c: Vertically down.
(iv) 0: No direction (center).
The gravitational field intensity at the center of a spherical shell is zero due to symmetry. However, because the mass distribution is only over a hemisphere in this case, there is no complete symmetry. Therefore, the resulting gravitational intensity would not be zero. For a hemispherical shell, the net gravitational force would be directed towards its flat surface due to the mass causing an imbalance.
Since the flat surface is horizontal and located above the center of the hemispherical shell:
(ii) b, Vertically up is the correct answer and indicates the direction of the gravitational field intensity at the center of the hemispherical shell.
For the above problem, the direction of the gravitational intensity at an arbitrary point $\mathrm{P}$ is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.
To help answer this question, let's first clarify what gravitational intensity (also called gravitational field strength) is. According to the chapter:
Gravitational intensity at a point is the force experienced by a unit mass placed at that point.
The direction of the gravitational intensity due to a mass is always directed towards the mass creating the gravitational field.
Given this, to find the correct direction of the gravitational intensity at an arbitrary point P, we need to know the distribution of mass around P.
Referring to Figure 7.7:
If point P is inside a spherical shell, the gravitational force at P is zero, hence no intensity direction.
If point P is outside a spherical shell, the direction of the gravitational force (and thus the intensity) is towards the center of the shell (along the line joining the center of the shell to point P).
For any other extended mass:
The direction of gravitational intensity at P is towards the center of mass of the object creating the field.
Therefore, without the specific diagram (d, e, f, g) to reference, we apply these principles:
Towards the center of mass of the object creating the gravitational field.
Given more specifics or a figure, the exact arrow can be identified accordingly.
A rocket is fired from the earth towards the sun. At what distance from the earth's centre is the gravitational force on the rocket zero ? Mass of the sun $=2 \times 10^{30} \mathrm{~kg}$, mass of the earth $=6 \times 10^{24} \mathrm{~kg}$. Neglect the effect of other planets etc. (orbital radius $=1.5 \times 10^{11} \mathrm{~m}$ ).
The solution to the equation $ \frac{R^2}{(1.5 \times 10^{11} - R)^2} = 3 \times 10^{-6} $ for ( R ) is given by:
[ R = \frac{150000000000 (1000 \sqrt{3} - 3)}{999997} ]
Calculating the numerical value,
[ R \approx \frac{150000000000 \times 170.08}{999997} \approx 255720 , \text{km} ]
Thus, the gravitational force on the rocket is zero at a distance of approximately 255,720 km from the Earth's center.
How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun is $1.5 \times 10^{8} \mathrm{~km}$.
The mass of the Sun (( M_{\text{S}} )) is approximately:
$$ M_{\text{S}} = 2.01 \times 10^{30} \text{ kg} $$
This value is derived using Kepler's third law with the given mean orbital radius of the Earth around the Sun and the gravitational constant.
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is $1.50 \times 10^{8} \mathrm{~km}$ away from the sun?
The semi-major axis of Saturn's orbit around the Sun is approximately: $$ \mathbf{1.43 \times 10^9 ; \text{km}} $$
So, Saturn is approximately 1.43 billion kilometers away from the Sun.
A body weighs $63 \mathrm{~N}$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ?
The gravitational force on the body at a height equal to half the radius of the Earth is 28 N.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed $250 \mathrm{~N}$ on the surface?
To determine the weight of a body halfway to the center of the Earth when it weighs $250 , \text{N}$ on the surface, we can use the fact that the gravitational force inside the Earth is directly proportional to the distance from the center.
Step-by-Step Solution:
Weight on the Surface:[ W_{\text{surface}} = 250 , \text{N} ]
Gravitational Force Inside the Earth:At a depth (d), the gravitational force (g(d)) is given by: [ g(d) = g \left(1 - \frac{d}{R_E}\right) ] where:
(d) is the depth from the surface
(R_E) is the radius of the Earth.
Halfway to the Center:Halfway to the center means (d = \frac{R_E}{2}).
Gravitational Acceleration at Halfway to the Center:[ g\left(\frac{R_E}{2}\right) = g \left(1 - \frac{\frac{R_E}{2}}{R_E}\right) = g \left(1 - \frac{1}{2}\right) = \frac{g}{2} ]
Weight at Halfway to the Center:The weight at halfway is proportional to the gravitational acceleration at that point. [ W_{\text{halfway}} = W_{\text{surface}} \cdot \frac{g_{\text{halfway}}}{g_{\text{surface}}} ]
Where ( g_{\text{halfway}} = \frac{g}{2} ).
Thus, [ W_{\text{halfway}} = 250 , \text{N} \cdot \frac{\frac{g}{2}}{g} = 250 , \text{N} \cdot \frac{1}{2} = 125 , \text{N} ]
Final Answer:
The weight of the body halfway to the center of the Earth would be 125 N.
A rocket is fired vertically with a speed of $5 \mathrm{~km} \mathrm{~s}^{-1}$ from the earth's surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth $=6.0 \times 10^{24} \mathrm{~kg}$; mean radius of the earth $=6.4 \times 10^{6} \mathrm{~m} ; G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$.
The distance ( R ) from the center of the Earth where the rocket will reach its highest point is computed to be approximately ( 6.4 \times 10^{6} , \text{m} ).
To find the height (( h )) above the Earth's surface:
$$ h = R - R_E $$
Since ( R \approx R_E ), it implies that the rocket will not go significantly far from the Earth’s surface before it starts falling back. However, due to the given precision, it indicates that the rocket barely lifts off and essentially returns quickly.
Thus, the height above the Earth's surface is extremely small, approximately zero, meaning the initial speed given is insufficient to gain any notable height.
The escape speed of a projectile on the earth's surface is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
The speed of the body far away from the Earth is:
$$ v_f = 31.68 \ \mathrm{km} \ \mathrm{s}^{-1} $$
This means that when the body is very far from the Earth (ignoring the presence of the sun and other planets), it will be moving at a speed of approximately 31.68 km/s.
A satellite orbits the earth at a height of $400 \mathrm{~km}$ above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite $=200 \mathrm{~kg}$; mass of the earth $=6.0 \times 10^{24} \mathrm{~kg}$; radius of the earth $=6.4 \times 10^{6} \mathrm{~m} ; G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$.
1. Gravitational Potential Energy (GPE):
The gravitational potential energy (U) at 400 km altitude is: [ U = -1.177 \times 10^{10} , \text{J} ]
2. Kinetic Energy (KE) Required to Escape:
The kinetic energy needed to achieve escape velocity (v_e) of (4472 , \text{m/s}) is: [ \text{KE} = 1.9998784 \times 10^9 , \text{J} ]
3. Total Energy Expended:
The total energy needed to move the satellite from its GPE at 400 km altitude to be free from Earth's gravitational influence is the sum of the magnitudes of GPE and KE: [ E_{\text{total}} = \left| U \right| + \text{KE} = 1.177 \times 10^{10} , \text{J} + 1.9998784 \times 10^9 , \text{J} ]
Therefore: [ E_{\text{total}} = 1.37698784 \times 10^{10} , \text{J} \approx 1.38 \times 10^{10} , \text{J} ]
Thus, approximately (1.38 \times 10^{10} , \text{J}) of energy must be expended to rocket the satellite out of Earth’s gravitational influence.
Two stars each of one solar mass ( $=2 \times 10^{30} \mathrm{~kg}$ ) are approaching each other for a head on collision. When they are a distance $10^{9} \mathrm{~km}$, their speeds are negligible. What is the speed with which they collide ? The radius of each star is $10^{4} \mathrm{~km}$. Assume the stars to remain undistorted until they collide. (Use the known value of $G$ ).
The speed ( v ) with which the two stars collide is approximately ( 1.29 \times 10^6 ) m/s.
*Initial separation:* ( r = 10^9 , \text{km} ) *Separation at collision:* ( 2 \times 10^4 , \text{km} ) *Change in potential energy:* ( \Delta U \approx -1.335 \times 10^{43} , \text{J} ) *Speed at collision:* ( v \approx 1.29 \times 10^6 , \text{m/s} )
Thus, the stars collide with a speed of approximately 1.29 million meters per second.
Two heavy spheres each of mass $100 \mathrm{~kg}$ and radius $0.10 \mathrm{~m}$ are placed $1.0 \mathrm{~m}$ apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
It appears there's difficulty in processing this query through the tool directly. Let's derive the results manually using the provided formulas.
Gravitational Force Between Two Spheres
The gravitational force ( F ) between two masses ( m_1 ) and ( m_2 ) separated by a distance ( r ) is given by: $$ F = G \frac{m_1 m_2}{r^2} $$
Using:
( G = 6.67 \times 10^{-11} , \mathrm{N , m^2 , kg^{-2}} )
( m_1 = 100 , \mathrm{kg} )
( m_2 = 100 , \mathrm{kg} )
( r = 1 , \mathrm{m} )
We get: $$ F = 6.67 \times 10^{-11} \frac{100 \times 100}{1^2} = 6.67 \times 10^{-11} \times 10000 = 6.67 \times 10^{-7} , \mathrm{N} $$
Gravitational Potential at Midpoint
The gravitational potential due to a mass ( M ) at a distance ( r ) is: $$ V = -G \frac{M}{r} $$
For two masses, the total potential at the midpoint will be the sum of potentials due to each mass:
Each mass ( M = 100 , \mathrm{kg} )
Distance to midpoint ( r = 0.5 , \mathrm{m} )
The potential from one mass: $$ V_1 = -G \frac{100 , \mathrm{kg}}{0.5 , \mathrm{m}} = -6.67 \times 10^{-11} \frac{100}{0.5} = -6.67 \times 10^{-11} \times 200 = -1.334 \times 10^{-8} , \mathrm{J , kg^{-1}} $$
Total potential: $$ V_{\text{total}} = 2 \times V_1 = 2 \times -1.334 \times 10^{-8} = -2.668 \times 10^{-8} , \mathrm{J , kg^{-1}} $$
Equilibrium and Stability
Equilibrium: The object at the midpoint is in equilibrium because the gravitational forces from both spheres cancel each other out.
Stability:
Unstable: If the object is displaced slightly, the gravitational pull from the nearer mass becomes stronger, pulling it further away from the midpoint.
Summary
Gravitational Force: ( 6.67 \times 10^{-7} , \mathrm{N} )
Gravitational Potential at Midpoint: ( -2.668 \times 10^{-8} , \mathrm{J , kg^{-1}} )
Equilibrium: Yes, but unstable.
Feel free to ask any further questions!
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Ask Chatterbot AINotes - Gravitation | Class 11 NCERT | Physics
Comprehensive Gravitation Class 11 Notes: Key Concepts and Important Laws
Introduction
Early Observations and Galileo’s Contribution
Gravitation, a fundamental force of nature, was first observed by ancient civilisations who noted that objects fall towards the Earth. Galileo Galilei, an Italian physicist, and astronomer, made significant contributions to our understanding of gravity. Through experiments involving rolling objects down inclined planes, he demonstrated that all objects, irrespective of their masses, accelerate towards the Earth at a constant rate.
The Nature of Gravitation
Gravitation explains why objects thrown upwards fall back to the ground and why climbing uphill is more strenuous than walking downhill. It is the force of attraction between any two masses.
Kepler’s Laws of Planetary Motion
Understanding Kepler’s Laws
Johannes Kepler derived three important laws from the meticulous observations of planetary motions made by Tycho Brahe.
Law of Orbits
- All planets move in elliptical orbits with the Sun at one of the focal points.
Law of Areas
- The line joining a planet and the Sun sweeps out equal areas during equal intervals of time.
Law of Periods
- The square of the period of revolution of a planet is proportional to the cube of the semi-major axis of its orbit.
Newton’s Universal Law of Gravitation
The Law Defined
Isaac Newton proposed the Universal Law of Gravitation, stating:
- Every mass attracts every other mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.
Mathematical Expression
For two masses ( m_1 ) and ( m_2 ) separated by a distance ( r ): [ F = G \frac{m_1 m_2}{r^2} ]
Understanding the Vector Form
The force vector ( \mathbf{F} ) acting on ( m_1 ) due to ( m_2 ) is: [ \mathbf{F} = -G \frac{m_1 m_2}{r^2} \hat{\mathbf{r}} ] where ( \hat{\mathbf{r}} ) is a unit vector from ( m_1 ) to ( m_2 ).
Gravitational Constant
Definition and Significance
The gravitational constant (G) is a proportionality constant in Newton's law of gravitation. Henry Cavendish first measured it.
Cavendish’s Experiment
In 1798, Cavendish used a torsion balance to measure ( G ), finding its value to be ( 6.67 \times 10^{-11} , \text{N} , \text{m}^2 , \text{kg}^{-2} ).
Gravitational Potential Energy
Concept of Potential Energy in Gravitational Field
Gravitational potential energy (U) is the energy possessed by an object due to its position in a gravitational field.
Formula and Calculation
The gravitational potential energy between two masses ( m_1 ) and ( m_2 ) separated by a distance ( r ) is: [ U = -G \frac{m_1 m_2}{r} ]
Potential vs. Potential Energy
Gravitational potential is the potential energy per unit mass. It is given by: [ V = -\frac{G M}{r} ]
Acceleration Due to Gravity
On the Surface of Earth
The acceleration due to gravity ( g ) at the Earth's surface is about ( 9.8 , \text{m/s}^2 ).
Above and Below the Surface of Earth
Variation with Altitude [ g(h) = g \left( 1 - \frac{2h}{R_E} \right) ]
Variation with Depth [ g(d) = g \left( 1 - \frac{d}{R_E} \right) ]
Escape Speed
Concept and Importance
Escape speed is the minimum speed needed for an object to break free from a planet's gravitational pull.
Derivation of Escape Speed
The escape speed ( v_e ) from Earth's surface can be derived as: [ v_e = \sqrt{2 g R_E} ] Numerically, this equals approximately ( 11.2 , \text{km/s} ).
Satellites
Natural vs. Artificial Satellites
Earth's natural satellite is the Moon. Artificial satellites are man-made objects launched into orbit for various purposes, including communication, weather monitoring, and space exploration.
Orbital Mechanics
Satellites in orbit are influenced by gravitational forces and follow Kepler's laws. The speed ( v ) of a satellite in a circular orbit is given by: [ v = \sqrt{\frac{G M_E}{R_E + h}} ]
Applications of Kepler’s Laws to Satellites
Kepler's third law applies to satellites revolving around Earth, helping in calculating orbital periods and distances.
graph TD
A[Kepler's Laws] --> B[Law of Orbits: Elliptical path]
A --> C[Law of Areas: Equal areas]
A --> D[Law of Periods: Orbital Period ~ Distance]
Practical Applications of Gravitation
Measuring Earth’s Mass
Earth's mass can be calculated using the formula: [ M_E = \frac{g R_E^2}{G} ]
Using Gravitational Concepts
Gravitational principles are crucial in space exploration, satellite technology, and estimating planetary masses.
Conclusion
Recap of Key Points
Understanding gravitation is essential in explaining various natural phenomena and technological applications. From the fall of an apple to the motion of planets, gravitation plays a vital role in our universe.
Importance of Gravity in Science
Gravitation forms the basis for theories in astrophysics, planetary science, and space exploration, making it a cornerstone in the scientific study of the cosmos.
Future Research in Gravitation
Research in gravitation continues to evolve, aiming to solve fundamental questions about the nature of the universe, such as the behaviour of gravity in extreme conditions like black holes and the unification with other fundamental forces.
By mastering these concepts, students will have a solid foundation in gravitation, one of the critical areas of study in physics.
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