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Question

For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $$R$$ is proportional to :


A
R1/2
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B
R3/2
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C
R1/2
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D
R0
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Solution

The correct option is D $${ R }^{ 0 }$$
Applying Newton's second law to a circular orbit, we have
$$\dfrac { m{ v }^{ 2 } }{ r } =\dfrac { 4{ \pi  }^{ 2 }rm }{ { T }^{ 2 } } =\dfrac { GMm }{ { r }^{ 2 } } $$
where, $$m$$ is the mass of satellite, and $$v$$ is the orbital speed.
$$T$$ is the time period
$$\therefore T=\dfrac { 2\pi { r }^{ { 3 }/{ 2 } } }{ \sqrt { GM }  } $$
For $$r\approx R$$
and $$M=\dfrac { 4 }{ 3 } \pi { R }^{ 3 }\rho $$ ($$\rho =$$ density of planet)
$$\because T=\sqrt { \dfrac { 3\pi  }{ \rho G }  } $$
i.e. $$T$$ is independent of $$R$$.

Physics

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