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Standard XII

Physics

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Two satellite...

Question

Two satellites, $A$ and $B$ , have masses $m$ and $2 m$ respectively. $A$ is in circular orbit of radius $R$ , and $B$ is in a circular orbit of radius $2 R$ around the earth. The ratio of their kinetic energies, $T_{A} / T_{B}$ is:

2

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\sqrt{\frac{1}{2}}

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1

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\frac{1}{2}

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Solution

The correct option is C $1$
Orbital velocity $V = \sqrt{\frac{G M e}{r}}$
$T_{A} = \frac{1}{2} m_{A} V_{A}^{2}$
$T_{B} = \frac{1}{2} m_{B} V_{B}^{2}$
$⟹ \frac{T_{A}}{T_{B}} = \frac{m \times \frac{G m}{R}}{2 m \times \frac{G m}{2 R}}$
$⟹ \frac{T_{A}}{T_{B}} = 1$

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Two satellites, A and B, have masses m and 2m respectively. A is in circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, TA/TB is:

The correct option is C 1Orbital velocity V=√GMerTA=12mAV2ATB=12mBV2B⟹TATB=m×GmR2m×Gm2R⟹TATB=1

Two satellites, $A$ and $B$ , have masses $m$ and $2 m$ respectively. $A$ is in circular orbit of radius $R$ , and $B$ is in a circular orbit of radius $2 R$ around the earth. The ratio of their kinetic energies, $T_{A} / T_{B}$ is:

The correct option is C $1$
Orbital velocity $V = \sqrt{\frac{G M e}{r}}$
$T_{A} = \frac{1}{2} m_{A} V_{A}^{2}$
$T_{B} = \frac{1}{2} m_{B} V_{B}^{2}$
$⟹ \frac{T_{A}}{T_{B}} = \frac{m \times \frac{G m}{R}}{2 m \times \frac{G m}{2 R}}$
$⟹ \frac{T_{A}}{T_{B}} = 1$