The potential energy of a body of mass $3 kg$ on the surface of a planet is $54$ joule. The escape velocity will be (in $\frac{m}{s}$ )

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Solution

Potential energy
$\frac{G M_{p} m}{R_{p}} = 54 \Rightarrow \frac{G M_{p}}{R_{p}} \times 3 = 54$
$\Rightarrow \frac{G M_{p}}{R_{p}} = 18$
Escape velocity
$v_{e} = \sqrt{\frac{2 G M_{p}}{R_{p}}} = \sqrt{2 \times 18} = 6 m s^{- 1}$

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The potential energy of a body of mass 3 kg on the surface of a planet is 54 joule. The escape velocity will be (in ms)

Potential energy GMpmRp=54⇒GMpRp×3=54 ⇒GMpRp=18 Escape velocity ve=√2GMpRp=√2×18=6 ms−1

The potential energy of a body of mass $3 kg$ on the surface of a planet is $54$ joule. The escape velocity will be (in $\frac{m}{s}$ )

Potential energy
$\frac{G M_{p} m}{R_{p}} = 54 \Rightarrow \frac{G M_{p}}{R_{p}} \times 3 = 54$
$\Rightarrow \frac{G M_{p}}{R_{p}} = 18$
Escape velocity
$v_{e} = \sqrt{\frac{2 G M_{p}}{R_{p}}} = \sqrt{2 \times 18} = 6 m s^{- 1}$