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's surface is 11 kms -1, the escape velocity from the surface of the planet would be

Question

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's surface is 11 kms -1, the escape velocity from the surface of the planet would be (A) 1.1 kms -1 (B) 11 kms -1 (C) 110 kms -1 (D) 0.11 kms -1

Tardigrade Admin · Accepted Answer

((ve)p/(ve)e)=(√(2 G Mp/Rp)/√(2 G Me)Re)=√(Mp/Me) × (Re/Rp) =√(10 Me/Me) × (Re/Re / 10)=10 ∴ (ve)p=10 ×(ve)e =10 × 11=110 km / s

Q. 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's surface is $11 km s^{- 1}$ , the escape velocity from the surface of the planet would be

$1.1 km s^{- 1}$

$11 km s^{- 1}$

$110 km s^{- 1}$

$0.11 km s^{- 1}$

$\frac{( v _{e} ) _{p}}{( v _{e} ) _{e}} = \frac{\frac{2 G M _{p}}{R _{p}}}{\frac{2 G M _{e}}{R _{e}}} = \frac{M _{p}}{M _{e}} \times \frac{R _{e}}{R _{p}}$
$= \frac{10 M _{e}}{M _{e}} \times \frac{R _{e}}{R _{e} /10} = 10$
$∴ (v_{e})_{p} = 10 \times (v_{e})_{e}$
$= 10 \times 11 = 110 km / s$

Q. 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's surface is 11kms−1, the escape velocity from the surface of the planet would be

1.1kms−1

11kms−1

110kms−1

0.11kms−1