The time period of an earth satellite close to the surface of the earth is $83$ minutes. The time period of another earth satellite in an orbit at a distance of the earth's radius from its surface will be

83

minutes

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83 \sqrt{8}

minutes

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664

minutes

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249

minutes

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Solution

The correct option is C $83$ minutes
For first earth satellite close to the surface of the earth,
$r_{1} = R$ , $T_{1} = 83$ min
Another earth satellite in an orbit at a distance of the earth radius from the surface,
$r_{2} = R$ , $T_{2} = ?$
The time period, $T_{2} = T_{1} (\frac{r_{2}}{r_{1}})^{3 / 2}$
$T_{2} = T_{1} (1)^{3 / 2}$
$T_{2} = T_{1} = 83$ min
The correct option is A.

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The time period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of the earth's radius from its surface will be

The correct option is C 83 minutesFor first earth satellite close to the surface of the earth,r1=R, T1=83 minAnother earth satellite in an orbit at a distance of the earth radius from the surface,r2=R, T2=? The time period, T2=T1(r2r1)3/2T2=T1(1)3/2T2=T1=83 minThe correct option is A.

The time period of an earth satellite close to the surface of the earth is $83$ minutes. The time period of another earth satellite in an orbit at a distance of the earth's radius from its surface will be