Question 281: Gravitation

Question

The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value then its new time period will become

Accepted Answer

Let initial separation between Earth and satellite be $r$ and its time period is $T = 24$ hours.
If the separation is decreased to one fourth, new separation $r' = \dfrac{r}{4}$.
From
NCERT
, the time period of a satellite in orbit at distance $r$ from the center of Earth is
$$ T = 2\pi \sqrt{\frac{r^3}{GM}} $$
where $G$ is gravitational constant and $M$ is mass of the Earth.
Let new time period be $T'$:
$$ T' = 2\pi \sqrt{\frac{(r')^3}{GM}} $$
Substitute $r' = \dfrac{r}{4}$:
$ T' = 2\pi \sqrt{ \frac{ \left( \frac{r}{4} ight)^3 }{ GM } } = 2\pi \sqrt{ \frac{ r^3 }{ GM } 	imes \frac{1}{4^3} } $
$ = 2\pi \sqrt{ \frac{ r^3 }{ GM } 	imes \frac{1}{64} } $
$ = 2\pi \sqrt{ \frac{ r^3 }{ GM } } 	imes \frac{1}{8} $
But $2\pi \sqrt{ \frac{ r^3 }{ GM } } = T = 24$ hours
So,
$ T' = T 	imes \frac{1}{8} = 24 	imes \frac{1}{8} = 3 	ext{ hours} $
Correct option: A) 3 hours

Answer

3 hours

Answer

6 hours

Answer

24 hours

Answer

12 hours

Gravitation

The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value then its new time period will become

Choose the correct answer: