Question

The ratio of escape velocity on the surface of the Moon to orbital velocity on the Earth is ___________. (Take $R_e$ = 64 × ${10}^5$ m , $R_m$ = 1.74 × ${10}^6$ m)

• A. 0.1
• B. 0.2
• C. 0.3
• D. 0.4

Answer 1

The correct option is C

0.3

Step 1, Given data

The radius of the earth $R_e$= 64 × ${10}^5$ m

The radius of the moon $R_m$= 1.74 × ${10}^6$m

Acceleration due to gravity on the moon is one-sixth of the earth

Step 2, Finding the ratio

We can write

$\frac{V_{em}}{V_{oe}}=\frac{\sqrt{2g_m{R}_m}}{\sqrt{g_e{R}_e}}$

Putting all the values

$\frac{V_{em}}{V_{oe}}=\sqrt{\frac{2\times g_m\times R_m}{6\times g_m\times R_e}}=\sqrt{\frac{1\times 1.74\times {10}^6}{3\times 64\times {10}^5}}=\sqrt{0.009\times {10}^1}$

After solving we get

$\frac{V_{em}}{V_{oe}}=0.3$

Therefore the ratio is 0.3.

Hence the correct option is (C).