What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is (3.1)2 ms-2 and whose radius is 8100 km?

Question

What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is (3.1)2 ms-2 and whose radius is 8100 km? (A) 2790 km-s-1 (B) 27.9 km-s-1 (C) (27.9/√5) km-1 (D) 27.9√5 km-s-1

Tardigrade Admin · Accepted Answer

Escape velocity ve=√2gR Given, g=(3.1)2ms-2, R = 8100km = 8 1 00 x 103 m. = ve=√2×(3.1)2×8100×103 ve=12.5×103ms-1 ⇒ ve=12.5 kms-1 ≈(27.9/√5)kms-1

Q. What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is $(3.1)^{2} m s^{- 2}$ and whose radius is 8100 km?

$2790 km - s^{- 1}$

$27.9 km - s^{- 1}$

$\frac{27.9}{5} k m^{- 1}$

$27.9 5 km - s^{- 1}$

Escape velocity $v_{e} = 2 g R$
Given, $g = (3.1)^{2} m s^{- 2},$
$R = 8100 km = 8100 x 1 0^{3} m .$ = $v_{e} = 2 \times (3.1)^{2} \times 8100 \times 1 0^{3}$
$v_{e} = 12.5 \times 1 0^{3} m s^{- 1}$
$\Rightarrow v_{e} = 12.5 km s^{- 1}$
$\approx \frac{27.9}{5} km s^{- 1}$

Q. What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is (3.1)2ms−2 and whose radius is 8100 km?

2790km−s−1

27.9km−s−1

5​27.9​km−1

27.95​km−s−1