Gravitation Ques 27
- A geostationary satellite orbits around the earth in a circular orbit of radius $36,000 \mathrm{~km}$. Then, the time period of a spy satellite orbiting a few hundred km above the earth’s surface $\left(R_{e}=6400 \mathrm{~km}\right)$ will approximately be
(2002)
(a) $1 / 2 \mathrm{~h}$
(b) $1 \mathrm{~h}$
(c) $2 \mathrm{~h}$
(d) $4 \mathrm{~h}$
Show Answer
Answer:
Correct Answer: 27.(c)
Solution:
Formula:
- Time period of a satellite very close to earth’s surface is 84.6 min. Time period increases as the distance of the satellite from the surface of earth increases. So, time period of spy satellite orbiting a few $100 \mathrm{~km}$ above the earth’s surface should be slightly greater than $84.6 \mathrm{~min}$. Therefore, the most appropriate option is (c) or $2 \mathrm{~h}$.
BETA
