What would be the escape velocity (in km/s) from a planet of mass
![What would be the escape velocity (in km/s) from a planet of mass](/img/relate-questions.png)
A. 44/3
B. 11/3
C. 33/2
D. 33/4
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
CONCEPT:
- Escape velocity: The minimum velocity required to escape the gravitational field of the earth is called escape velocity.
- It is denoted by Ve.
- The escape velocity on earth is given by:
\({V_e} = \sqrt {\frac{{2\;G\;M_e}}{R_e}} \)
CALCULATION:
Given - Mass of a planet (Mp) = 1/4th of the mass of the earth (Me), the radius of a planet (rp) = 1/9 th the radius of the earth (re) and escape velocity from the earth (Ve) = 11 km/s
- The escape velocity on the earth is given by:
\(\Rightarrow {V_e} = \sqrt {\frac{{2\;G\;M_e}}{r_e}} \) --------- (1)
- The escape velocity on the planet is given by:
\(\Rightarrow {V_p} = \sqrt {\frac{{2\;G\;M_p}}{r_p}} =\sqrt {\frac{{2\;G\;\frac{M_e}{4}}}{\frac{r_e}{9}}}\)
\(\Rightarrow {V_p} = \sqrt {\frac{{2\;(9GM_e)}}{4e_e}}=\frac{3}{2}\sqrt {\frac{{2\;G\;M_e}}{r_e}} \) --------- (2)
On dividing equation 1 and 2, we get
\(\Rightarrow \frac{V_e}{V_p} =\frac{2}{3}\)
\(\Rightarrow V_p=\frac{3}{2}V_e=\frac{33}{2}\, km/s\)