Spatial Information Technology - Class 12 Geography - Chapter 4 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Spatial Information Technology | Practical Work in Geography | Geography | Class 12
A geosynchronous satellite is a satellite that orbits the Earth with an orbital period of 24 hours, thus matching the period of the Earth's rotational motion. A special class of geosynchronous satellites is a geostationary satellite. A geostationary satellite orbits the Earth in 24 hours along an orbital path that is parallel to an imaginary plane drawn through the Earth's equator. Such a satellite appears permanently fixed above the same location on the Earth. If a geostationary satellite wishes to orbit the Earth in 24 hours ($86400$ s), then how high above the Earth's surface must it be located? (Given: $M_{\text {Earth}} = 5.98 \times 10^{24}$ kg, $R_{\text {Earth}} = 6.37 \times 10^{6}$ m)
A $10.1 \times 10^{7}$ m
B $6.38 \times 10^{7}$ m
C $3.59 \times 10^{7}$ m
D $4.56 \times 10^{7}$ m
The correct answer is C: $$ 3.59 \times 10^{7} \mathrm{~m} $$
Given:
Orbital period $T = 86400 \text{ s}$
Mass of Earth $M_{\text{Earth}} = 5.98 \times 10^{24} \text{ kg}$
Radius of Earth $R_{\text{Earth}} = 6.37 \times 10^{6} \text{ m}$
Gravitational constant $G = 6.673 \times 10^{-11} \text{ Nm}^{2} / \text{kg}^{2}$
Using the formula for the period of a satellite in a circular orbit: $$ \frac{T^{2}}{R^{3}} = \frac{4 \pi^{2}}{GM_{\text{central}}} $$ we can rearrange it to solve for the cube of the radius $R$ of the orbit: $$ R^{3} = \frac{G M_{\text{central}} T^{2}}{4 \pi^{2}} $$ Substituting the given values: $$ R^{3} = \frac{(6.673 \times 10^{-11} \text{Nm}^{2}/\text{kg}^{2})(5.98 \times 10^{24} \text{kg})(86400 \text{s})^{2}}{4 \pi^{2}} $$ $$ R^{3} = 7.54 \times 10^{22} \text{m}^3 $$ Taking the cube root gives us the distance from the center of Earth to the satellite: $$ R = 4.23 \times 10^{7} \text{m} $$ To find the height above Earth's surface, subtract Earth's radius from the total orbital radius: $$ \text{Height} = R - R_{\text{Earth}} = 4.23 \times 10^{7} \text{m} - 6.37 \times 10^{6} \text{m} $$ $$ \text{Height} = 3.59 \times 10^{7} \text{m} $$ Thus, the required height of the geostationary satellite is $3.59 \times 10^{7} \text{m}$ above the Earth's surface.
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Choose the right answer from the four alternatives given below :
(i) The spatial data are characterised by the following forms of appearance :
(a) Positional (b) Linear (c) Areal (d) All the above forms
(ii) Which one of the following operations requires analysis module software?
(a) Data storage (b) Data display (c) Data output (d) Buffering
(iii) Which one of the following is disadvantage of Raster data format ?
(a) Simple data structure
(b) Easy and efficient overlaying
(c) Compatible with remote sensing imagery
(d) Difficult network analysis
(iv) Which one of the following is an advantage of Vector data format ?
(a) Complex data structure
(b) Difficult overlay operations
(c) Lack of compatibility with remote sensing data
(d) Compact data structure
(v) Urban change detection is effectively undertaken in GIS core using:
(a) Overlay operations
(b) Proximity analysis
(c) Network analysis
(d) Buffering
(i) (d) All the above forms
(ii) (d) Buffering
(iii) (d) Difficult network analysis
(iv) (d) Compact data structure
(v) (a) Overlay operations
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Write an explanatory account of the sequence of activities involved in GIS related work. (125 words)
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