A geo-stationary satellite is orbiting the earth at a height of 6 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is

10 hr

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(- 6 / \sqrt{2}) h r

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6 hr

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6 \sqrt{2} h r

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Solution

The correct option is D $6 \sqrt{2} h r$
From Kepler’s Law

$T^{2} \propto r^{3}$

${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{r_{1}}{r_{2}})}^{3}$

$\frac{24}{T_{2}} = {(\frac{7 R}{3.5 R})}^{\frac{3}{2}}$

$T_{2} = \frac{12}{\sqrt{2}}$

$T_{2} = 6 \sqrt{2} H o u r s$

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A geo-stationary satellite is orbiting the earth at a height of 6 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is

The correct option is D 6√2hrFrom Kepler’s Law T2∝r3 (T1T2)2=(r1r2)3 24T2=(7R3.5R)32 T2=12√2 T2=6√2Hours

The correct option is D $6 \sqrt{2} h r$
From Kepler’s Law

$T^{2} \propto r^{3}$

${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{r_{1}}{r_{2}})}^{3}$

$\frac{24}{T_{2}} = {(\frac{7 R}{3.5 R})}^{\frac{3}{2}}$

$T_{2} = \frac{12}{\sqrt{2}}$

$T_{2} = 6 \sqrt{2} H o u r s$