The period of revolution of an earth satellite close to the surface of the earth is 90 min. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be :

Question

The period of revolution of an earth satellite close to the surface of the earth is 90 min. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be : (A) 90 min (B)  90× √8 min  (C) 270 min (D) 720 min

Tardigrade Admin · Accepted Answer

The period of revolution of a satellite depends only upon its height above earths surface, for a satellite very close to the surface of the earth, period of revolution is

T_{1} = 2 π \frac{R}{g}

.... (i)
For a height h above the earths surface,

T_{2} = \frac{2 π}{R} \frac{( R + h ) ^{3}}{g}

... (ii)
Given,

h = 3 R

∴ T_{2} = \frac{2 π}{R} = \frac{( R + 3 R ) ^{3}}{g}

= 2 π \times 8 \frac{R}{g}

... (iii)
From Eqs. (i) and (iii), we have

\frac{T _{1}}{T _{2}} = \frac{1}{8}

\Rightarrow T_{2} = 90 \times 8 = 720 min

Q. The period of revolution of an earth satellite close to the surface of the earth is $90 min$ . The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be :

90 min

$90 \times 8 min$

270 min

720 min

Q. The period of revolution of an earth satellite close to the surface of the earth is 90min. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be :

90 min

90×8​min

270 min

720 min

Q. The period of revolution of an earth satellite close to the surface of the earth is $90 min$ . The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be :

$90 \times 8 min$