A planet in a distant solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11 km/s$ , the escape velocity from the surface of the planet would be

1.1 km/s

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11 km/s

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110 km/s

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0.11 km/s

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Solution

The correct option is C $110 km/s$
Mass of planet, $M_{p} = 10 M_{e}$ , where $M_{e}$ is mass of earth.
Radius of planet, $R_{P} = \frac{R_{e}}{10}$ , Where $R_{e}$ is radius of earth.
Escape speed is given by, $V_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}} = 11 km/s$
For planet, $V_{P} = \sqrt{\frac{2 G \times M_{P}}{R_{P}}} = \sqrt{\frac{100 \times 2 G M_{e}}{R_{e}}}$
$= 10 \times V_{e}$
$= 10 \times 11 = 110 Km/s$

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A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 km/s, the escape velocity from the surface of the planet would be

The correct option is C 110 km/sMass of planet, Mp=10Me, where Me is mass of earth. Radius of planet, RP=Re10, Where Re is radius of earth. Escape speed is given by, Ve=√2GMeRe=11 km/s For planet, VP=√2G×MPRP=√100×2GMeRe =10×Ve =10×11=110 Km/s

A planet in a distant solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11 km/s$ , the escape velocity from the surface of the planet would be