The ratio of the time periods of two satellites at heights 3600 km, 13,600 km from the surface of the earth is

1 : 2.828

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2.828 : 1

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2 : 828 : 2

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1 : 1

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Solution

The correct option is A 1 : 2.828
Kepler’s law-

$T α {(R + h)}^{3 / 2}$

$\frac{T_{1}}{T_{2}} = {(\frac{R + h_{1}}{R + h_{2}})}^{3 / 2} = {(\frac{10, 000}{20, 000})}^{3 / 2}$

$= \frac{1}{2.828}$

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The ratio of the time periods of two satellites at heights 3600 km, 13,600 km from the surface of the earth is

The correct option is A 1 : 2.828Kepler’s law- Tα(R+h)3/2 T1T2=(R+h1R+h2)3/2=(10,00020,000)3/2 =12.828

The correct option is A 1 : 2.828
Kepler’s law-

$T α {(R + h)}^{3 / 2}$

$\frac{T_{1}}{T_{2}} = {(\frac{R + h_{1}}{R + h_{2}})}^{3 / 2} = {(\frac{10, 000}{20, 000})}^{3 / 2}$

$= \frac{1}{2.828}$