Question

A satellite having mass 1/4 of the earth mass what will be the time taken comeplete on (revolution) orbit above the earth's surface at height of 35780km?

Answer 1

Answer:

maybe 1 to 2 days

Explanation:

we can't decide exactly

Answer 2

The time period of revolution of satellite is 38.35 hr.

Explanation:

Given that,

Mass of satellite $M_{s}=\dfrac{1}{4}M_{e}$

Height of the satellite= 35780 km

We need to calculate the time taken

Time taken by the satellite to revolved around the earth's orbit

$T=\dfrac{2\pi (R+h)}{v_{c}}$...(I)

We know that,

$v_{c}=\sqrt{\dfrac{GM_{e}}{R+h}}$

Now, Put the value of v in equation (I)

$T=\dfrac{2\pi (R+h)}{\sqrt{\dfrac{GM_{e}}{R+h}}}$

$T=\dfrac{2\pi (R+h)\times\sqrt{R+h}}{\sqrt{GM_{e}}}$...(II)

$T\propto\dfrac{1}{\sqrt{M_{e}}}$

Put the value in the equation (II)

$T=\dfrac{2\pi(6.4\times10^{5}+35780\times10^{3})\times\sqrt{6.4\times10^{5}+35780\times10^{3}}}{\sqrt{6.67\times10^{-11}\times\dfrac{6\times10^{24}}{4}}}$

$T=138064.24\ sec$

$T=38.35\ hr$

Hence, The time period of revolution of satellite is 38.35 hr.

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Topic : time period

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