# What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 x 10^24 kg and radius of the earth is 6.4 x 10^6 m.

## Question

What is the magnitude of the gravitational force between the earth and a $1 \mathrm{~kg}$ object on its surface? (Mass of the earth is $6 \times 10^{24} \mathrm{~kg}$ and radius of the earth is $6.4 \times 10^{6} \mathrm{~m}$.

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## Answer

The gravitational force between two objects, according to Newton's law of universal gravitation, is given by the formula:

$$F = \frac{G m_1 m_2}{r^2}$$

where $F$ is the gravitational force, $G$ is the gravitational constant ($6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2$), $m_1$ is the mass of the first object, $m_2$ is the mass of the second object, and $r$ is the distance between the centers of the two objects.

Given:

Mass of the Earth ($m_1$): $6 \times 10^{24} \, \text{kg}$

Mass of the object ($m_2$): $1 \, \text{kg}$

Radius of the Earth ($r$): $6.4 \times 10^{6} \, \text{m}$

Plugging in these values:

$$F = \frac{(6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2)(6 \times 10^{24} \, \text{kg})(1 \, \text{kg})}{(6.4 \times 10^{6} \, \text{m})^2}$$

Solving this, we can find the magnitude of the gravitational force between the Earth and a $1 \, \text{kg}$ object on its surface. The magnitude of the gravitational force between the Earth and a $1 \, \text{kg}$ object on its surface is approximately $9.776 \, \text{N}$ (newtons).

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