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Standard X

Physics

Need and Importance of Space Mission

A geostationa...

Question

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

6 \sqrt{2} h r

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6 h r

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5 h r

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10 h r

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Solution

The correct option is A $6 \sqrt{2} h r$
Distance of satellite from center of the earth are $7 R$ and $3.5 R$ respectively.
Time period of geostationary satellite is $24 h r$
$\frac{T_{2}}{T_{1}} = {(\frac{R_{2}}{R_{1}})}^{3 / 2}$
$T_{2} = 24 {(\frac{3.5 R}{7 R})}^{3 / 2} = 6 \sqrt{2} h r$

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

The correct option is A 6√2 hrDistance of satellite from center of the earth are 7R and 3.5R respectively.Time period of geostationary satellite is 24hr T2T1=(R2R1)3/2 T2=24(3.5R7R)3/2=6√2hr

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

The correct option is A $6 \sqrt{2} h r$
Distance of satellite from center of the earth are $7 R$ and $3.5 R$ respectively.
Time period of geostationary satellite is $24 h r$
$\frac{T_{2}}{T_{1}} = {(\frac{R_{2}}{R_{1}})}^{3 / 2}$
$T_{2} = 24 {(\frac{3.5 R}{7 R})}^{3 / 2} = 6 \sqrt{2} h r$