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
B
R3/2
C
R1/2
D
R0

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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