Question

What is the escape velocity on the surface of the
moon given the mass and radius of the moon to be
7.34x1022 kg and 1.74x10 m respectively.
(1) 2 m/s
(2) 2 km/s
(3) 2.37 m/s (4) 2.37 km/s​

Answer 1

Answer:

Escape velocity of moon is 2.38km/s

Explanation:

$ve = \sqrt{\frac{2GM}{r} }$

Given

Mass = 7.34 x 10^22Kg

Radius = 1.74x10^6m

Ve = √(6.67x10^-24 x 7.34x10^22)/1.74x10^6

Ve = 2370m/s

Ve = 2.37Km/s

Answer 2

$\pink{ \boxed{\boxed{\begin{array}{cc} \maltese \: \rm{\bf \: \: given }\: \\ \sf \rightarrow \: \: mass \: of \: moon \: \:M = 7.34 \times {10}^{22} \: kg \\ \sf \rightarrow \:radius \: o f \: moom \: \: R = 1.74 \times {10}^{6} \: m \\ \\ \sf \rightarrow \:escape \: velocity \: of \: moon \: \: v_m = \: to \: find \\ \\ \red{ \bf \: we \: know \: that \: } \\ \bf \: v_m = \sqrt{ \frac{2GM}{R} } \\ = \sqrt{ \frac{2 \times 6.673 \times {10}^{ - 11} \times 7.34 \times {10}^{22} }{1.74 \times {10}^{6} } } \\ = 2372.8 \: m {s}^{ - 1} \\ = 2.37 \: km {s}^{ - 1} \\ \\ \blue{ \boxed{\therefore \: \bf \: v_m = 2.37 \: km {s}^{ - 1}}} \end{array}}}}$