A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth; R being the radius of the earth. What will be the time period of another satellite at a height 2.5 R from the surface of the earth?

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A geostationary satellite is orbiting the earth at a height of 6 R from the earth's surface (R is the earth's radius). What is the period of rotation of another satellite at a height of 2.5 R from the earth's surface.

The correct option is B 6√2 hoursT=2π√r3GM(T1T2)2=(r1r2)3=(6R+R2.5R+R)3=8T2=T1√8=24√8=6√2 hr

A geostationary satellite is orbiting the earth at a height of $6$ R from the earth's surface (R is the earth's radius). What is the period of rotation of another satellite at a height of $2.5$ R from the earth's surface.

The correct option is B $6 \sqrt{2}$ hours
$T = 2 π \sqrt{\frac{r^{3}}{G M}} {(\frac{T_{1}}{T_{2}})}^{2} {= (\frac{r_{1}}{r_{2}})}^{3} = {(\frac{6 R + R}{2.5 R + R})}^{3} = 8 T_{2} = \frac{T_{1}}{\sqrt{8}} = \frac{24}{\sqrt{8}} = 6 \sqrt{2} h r$