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Standard XII

Physics

Variation of G Due to Height and Depth

A satellite i...

Question

A satellite is revolving in a circular orbit at a distance of $2620 k m$ from the surface of the earth. The time period of revolution of the satellite is
(Radius of the earth $= 6380 k m$ , mass of the earth $= 6 \times 10^{24} k g$ , $G = 6.67 \times 10^{- 11} N - m^{2} / k g^{2}$ )

2.35 h o u r s

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23.5 h o u r s

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3.25 h o u r s

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32.5 h o u r s

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Solution

The correct option is A $2.35 h o u r s$
Orbital velocity, $v_{0} = \sqrt{\frac{G M_{e}}{R_{e} + h}}$
$= \sqrt{\frac{(6.67 \times 10^{- 11}) (6 \times 10^{24})}{6380 + 2620}}$
$= 6.67 k m / s e c$
Time-period of revolution,
$T = 2 π \frac{(R_{e} + h)}{v_{0}}$
$= \frac{2 \times 3.14 \times (6380 + 2620)}{6.67 \times 10^{3}}$
$= 8474 s e c = 2.35 h o u r s$

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A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The time period of revolution of the satellite is (Radius of the earth =6380 km, mass of the earth =6×1024kg, G=6.67×10−11N−m2/kg2)

The correct option is A 2.35 hoursOrbital velocity, v0=√GMeRe+h=√(6.67×10−11)(6×1024)6380+2620=6.67km/secTime-period of revolution,T=2π(Re+h)v0=2×3.14×(6380+2620)6.67×103=8474sec=2.35hours

A satellite is revolving in a circular orbit at a distance of $2620 k m$ from the surface of the earth. The time period of revolution of the satellite is
(Radius of the earth $= 6380 k m$ , mass of the earth $= 6 \times 10^{24} k g$ , $G = 6.67 \times 10^{- 11} N - m^{2} / k g^{2}$ )