**Solve the problem.**

How much time a satellite in an orbit at height 35780 km above earth's surface would take, if the mass of the earth would have been four times its original mass?

#### Solution

Given:

Height of the satellite, h = 35780 km

Let the original mass of Earth be M. Then its new mass will be 4M.

Time taken by the satellite to revolved around the Earth's orbit is given as

\[T = \frac{2\pi R}{v_c}\]

Now, v_{c }is given as

\[T = \frac{2\pi(R + h)}{\sqrt{\frac{GM}{R + h}}} = \frac{2\pi(R + h)\sqrt{R + h}}{\sqrt{GM}} . . . . . (i)\]

\[ \Rightarrow T \propto \frac{1}{\sqrt{M}} . . . . . (ii)\]

Thus from equation (ii) we see that when the mass of the Earth becomes 4 times, the time period of revolution of satellite should be halved.

i.e. \[T_{4M} = \frac{T}{2}\] .....(iii)

Now, h = 35780 km

M = \[6 \times {10}^{24} kg\]

^{-11}N m

^{2}/kg

^{2}

Putting the values of h, M and R in first, we get

T = 24 h

Using (iii), we get