Question 1:
moderateThe work done to raise a mass $m$ from the surface of the earth to a height $h$, which is equal to the radius of the earth, is:
(2019)
Work done $W = \Delta U = \frac{mgh}{1+h/R}$. Substituting $h = R$, we get $W = \frac{mgR}{1+1} = \frac{1}{2}mgR$.
Question 2:
moderateA body of mass ‘$m$’ taken from the earth’s surface to the height equal to twice the radius ($R$) of the earth. The change in potential energy of body will be:
(2013)
Change in potential energy $\Delta U = \frac{mgh}{1+h/R}$. For $h = 2R$, $\Delta U = \frac{mg(2R)}{1+2} = \frac{2}{3}mgR$.
Question 3:
easyA body of mass $m$ is placed on earth surface which is taken from earth surface to a height of $h = 3R$ then change in gravitational potential energy is:
(2003)
Change in potential energy $\Delta U = \frac{mgh}{1+h/R}$. Given $h = 3R$, $\Delta U = \frac{mg(3R)}{1+3} = \frac{3}{4}mgR$.
Question 4:
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 5:
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 6:
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 7:
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, $v'_e = 2 \times 11.2\text{ km/s} = 22.4\text{ km/s}$.
Question 8:
easyThe escape velocity from earth is $11.2\text{ km/s}$. If a body is to be projected in a direction making an angle $45^{\circ}$ to the vertical, then the escape velocity is:
(1993)
Escape velocity is a scalar quantity and is independent of the direction or angle of projection. It depends only on the mass and radius of the planet, remaining $11.2\text{ km/s}$.
Question 9:
easyThe satellite of mass $m$ is orbiting around the earth in a circular orbit with a velocity $v$. What will be its total energy?
(1991)
Kinetic energy of the satellite is $K = \frac{1}{2}mv^2$ and potential energy is $U = -mv^2$. Total energy $E = K + U = \frac{1}{2}mv^2 - mv^2 = -\frac{1}{2}mv^2$.
Question 10:
easyFor a satellite escape velocity is $11\text{ km/s}$. If the satellite is launched at an angle of $60^{\circ}$ with the vertical, then escape velocity will be:
(1989)
The escape velocity formula depends solely on the gravitational potential and mass/radius of the body. It is independent of the angle at which the object is launched.