Assuming g(moon) = ( (1/6)) gearth and D(moon) = ( (1/4)) Dearth where g and D are. the acceleration due to gravity and diameter respectively, the escape velocity from the moon is

Question

Assuming g(moon) = ( (1/6)) gearth and D(moon) = ( (1/4)) Dearth where g and D are. the acceleration due to gravity and diameter respectively, the escape velocity from the moon is (A) (11.2/24) km s-1 (B)  11.2 × √24 k m s-1 (C) (11.2/√24) km s-1 (D)  11.2 × 24 k m s-1

Tardigrade Admin · Accepted Answer

Escape velocity of any planet is given by,

v = 2 g R

where g = acceleration due to gravity
R = radius of the planet
Here,

g_{(m oo n)} = \frac{1}{6} g_{e a r t h}

D_{(m oo n)} = \frac{1}{4} D_{e a r t h}

or,

R_{(m oo n)} = \frac{1}{4} R_{e a r t h}

∴ v_{m oo n} = 2 \times \frac{1}{6} g_{e a r t h} \times \frac{1}{4} R_{e a r t h}

= \frac{1}{24} 2_{g_{e a r t h}} R_{e a r h} = \frac{v _{e a r t h}}{24}

= \frac{11.2}{24} km s^{- 1}

Q. Assuming $g_{(m oo n)} = (\frac{1}{6}) g_{e a r t h}$ and $D_{(m oo n)} = (\frac{1}{4}) D_{e a r t h}$ where g and D are. the acceleration due to gravity and diameter respectively, the escape velocity from the moon is

$\frac{11.2}{24} km s^{- 1}$

$11.2 \times 24 km s^{- 1}$

$\frac{11.2}{24} km s^{- 1}$

$11.2 \times 24 km s^{- 1}$

Q. Assuming g(moon)​=(61​)gearth​ and D(moon)​=(41​)Dearth​ where g and D are. the acceleration due to gravity and diameter respectively, the escape velocity from the moon is

2411.2​kms−1

11.2×24​kms−1

24​11.2​kms−1

11.2×24kms−1

Q. Assuming $g_{(m oo n)} = (\frac{1}{6}) g_{e a r t h}$ and $D_{(m oo n)} = (\frac{1}{4}) D_{e a r t h}$ where g and D are. the acceleration due to gravity and diameter respectively, the escape velocity from the moon is

$\frac{11.2}{24} km s^{- 1}$

$11.2 \times 24 km s^{- 1}$

$\frac{11.2}{24} km s^{- 1}$

$11.2 \times 24 km s^{- 1}$