Question 11:
moderateThe escape velocity from the Earth’s surface is $v$. The escape velocity from the surface of another planet having a radius four times that of Earth and same mass density is:
(2021)
Escape velocity $v = R \sqrt{\frac{8}{3} \pi G \rho}$. Since $v \propto R$ for constant density, a planet with 4 times the radius will have $v' = 4v$.
Question 12:
moderateA black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass $= 5.98 \times 10^{24}\text{ kg}$) have to be compressed to be a black hole?
(2014)
For a black hole, the escape velocity equals the speed of light $c$, leading to the radius formula $R = \frac{2GM}{c^2}$. Substituting the gravitational constant, mass of earth, and speed of light gives $R \approx 10^{-2}\text{ m}$. Thus, option A is correct.
Question 13:
moderateFor a planet having mass equal to mass of the earth but radius is one fourth of radius of the earth, Then escape velocity for this planet will be:
(2000)
Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$. Since mass is constant and radius becomes one-fourth, $v_e$ increases by a factor of $\sqrt{4} = 2$. Thus, $v_p = 2 \times 11.2\text{ km/s} = 22.4\text{ km/s}$. Option B is correct.
Question 14:
moderateThe escape velocity of a body on the surface of the earth is $11.2\text{ km/s}$. If the earth’s mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes:
(1997)
Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$. When mass becomes $2M$ and radius becomes $R/2$, $v'_e = \sqrt{\frac{2G(2M)}{R/2}} = 2v_e$. Thus, $v'_e = 2 \times 11.2\text{ km/s} = 22.4\text{ km/s}$.