The time period of a geo-stationary satellite is 24 ~h, at a height 6 R_E-R_E is the radius of earth) from surface of earth. The time period of…
NEET UG 2019 — Physics Mechanics
2019mcqeasy
The time period of a geo-stationary satellite is 24h, at a height 6RE−RE is the radius of earth) from surface of earth. The time period of another satellite whose height is 2.5RE from surface will be
Official previous-year question
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
A
62h
B
122h
C
2.524h
D
2.512h
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Solution
From Kepler's third law, the time period of revolution of satellite around earth is T2∝r3 or T∝r3/2 where, r is the radius of satellite's orbit. Here, r1=6RE+RE,T1=24hr2=2.5RE+RE,T2= ? where RE= radius of earth So, from Eq. (i), we get T2T1T224⇒T2=(r2r1)3/2=(2.5RE+RE6RE+RE)3/2=(3.57)3/2=(2)3/224=2224=212=62h
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