A geostationary satellite orbits around the earth in a circular orbit of radius $36000 k m$ . Then, the period of spy satellite orbiting three hundred kilometers above the earth's surface ( $R_{e a r t h} = 6400 k m$ ) will become :

(1 / 2) h r

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1.5 h r

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2 h r

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4 h r

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Solution

The correct option is C $2 h r$
$T_{1} =$ 24 hrs ; $R_{1} = 36000$ km
$R_{2} = 6700$ km
$T^{2} \propto R^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{R_{2}}{R_{1}})}^{3}$
${(\frac{T_{2}}{24})}^{2} = {(\frac{6700}{36000})}^{3}$
$T_{2} = 1.92 h r s$

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A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the period of spy satellite orbiting three hundred kilometers above the earth's surface (Rearth=6400 km) will become :

The correct option is C 2hrT1= 24 hrs ; R1=36000 kmR2=6700 km T2∝R3(T2T1)2=(R2R1)3(T224)2=(670036000)3T2=1.92 hrs

A geostationary satellite orbits around the earth in a circular orbit of radius $36000 k m$ . Then, the period of spy satellite orbiting three hundred kilometers above the earth's surface ( $R_{e a r t h} = 6400 k m$ ) will become :

The correct option is C $2 h r$
$T_{1} =$ 24 hrs ; $R_{1} = 36000$ km
$R_{2} = 6700$ km
$T^{2} \propto R^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{R_{2}}{R_{1}})}^{3}$
${(\frac{T_{2}}{24})}^{2} = {(\frac{6700}{36000})}^{3}$
$T_{2} = 1.92 h r s$