JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 10)
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational
field of the planet acts on the spaceship, what will be the number of complete revolutions made by the
spaceship in 24 hours around the planet?
[Given ; Mass of planet = 8 × 1022 kg, Radius of planet = 2 × 106 m, Gravitational constant G = 6.67 × 10–11 Nm2 /kg2]
[Given ; Mass of planet = 8 × 1022 kg, Radius of planet = 2 × 106 m, Gravitational constant G = 6.67 × 10–11 Nm2 /kg2]
13
9
17
11
Explanation
$${{m{V^2}} \over r} = {{GMm} \over {{r^2}}}$$
$$V = \sqrt {{{GM} \over r}} $$
$$n = {{VT} \over {2\pi r}} = \sqrt {{{GM} \over r}} {T \over {2\pi r}}$$
$$ = \left( {\sqrt {{{GM} \over {{r^3}}}} } \right) \times {T \over {2\pi }} = \sqrt {{{6.67 \times {{10}^{ - 11}} \times 8 \times {{10}^{22}}} \over {{{\left( {202 \times {{10}^4}} \right)}^3}}}} \times {T \over {2\pi }}$$
$$ = {{24 \times 3600} \over {2 \times 3.14}}\sqrt {{{6.67 \times 8 \times {{10}^{11}}} \over {{{\left( {202} \right)}^3} \times {{10}^{12}}}}} $$
$$ = {{24 \times 3600} \over {2 \times 3.14 \times 1242.8}} = {{24 \times 3600} \over {74.51}} = 11$$
$$V = \sqrt {{{GM} \over r}} $$
$$n = {{VT} \over {2\pi r}} = \sqrt {{{GM} \over r}} {T \over {2\pi r}}$$
$$ = \left( {\sqrt {{{GM} \over {{r^3}}}} } \right) \times {T \over {2\pi }} = \sqrt {{{6.67 \times {{10}^{ - 11}} \times 8 \times {{10}^{22}}} \over {{{\left( {202 \times {{10}^4}} \right)}^3}}}} \times {T \over {2\pi }}$$
$$ = {{24 \times 3600} \over {2 \times 3.14}}\sqrt {{{6.67 \times 8 \times {{10}^{11}}} \over {{{\left( {202} \right)}^3} \times {{10}^{12}}}}} $$
$$ = {{24 \times 3600} \over {2 \times 3.14 \times 1242.8}} = {{24 \times 3600} \over {74.51}} = 11$$
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