A geostationary satellite is orbiting the earth at a height 6R above its surface, where R is the radius of the earth. What is the time period of another satellite at a height of 2.5R from the surface of earth in hours?

Question

A geostationary satellite is orbiting the earth at a height 6R above its surface, where R is the radius of the earth. What is the time period of another satellite at a height of 2.5R from the surface of earth in hours? (A) 3√2h (B) 1.5√2h (C) 6√2h (D) 12√2h

Tardigrade Admin · Accepted Answer

Here, r1=6R+R=7R; T1 = 24 hours; r2-2.5R + R = 3.5R; T2= 1 T2=[(R2/R1)]3 / 2× T2=24[(3.5 R/7 R)]3 / 2=(24/2 √2)=6√2h h

Q. A geostationary satellite is orbiting the earth at a height $6 R$ above its surface, where $R$ is the radius of the earth. What is the time period of another satellite at a height of $2.5 R$ from the surface of earth in hours?

$32 h$

$1.5 2 h$

$62 h$

$122 h$

Here, $r_{1} = 6 R + R = 7 R; T_{1} = 24 h o u rs; r_{2} - 2.5 R + R = 3.5 R; T_{2} = 1$
$T_{2} = [\frac{R _{2}}{R _{1}}]^{3/2} \times T_{2} = 24 [\frac{3.5 R}{7 R}]^{3/2} = \frac{24}{2 2} = 62 h$ h

Q. A geostationary satellite is orbiting the earth at a height 6R above its surface, where R is the radius of the earth. What is the time period of another satellite at a height of 2.5R from the surface of earth in hours?

32​h

1.52​h

62​h

122​h