A 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 $h = 2R$, $\Delta U = \frac{mg(2R)}{1+2} = \frac{2}{3}mgR$.
A particle of mass $M$ is situated at the center of a spherical shell of same mass and radius $a$. The magnitude of the gravitational potential at a point situated at $\frac{a}{2}$ distance from the center, will be:
(2011 Mains)
Potential at distance $a/2$ is $V = V_{\text{shell}} + V_{\text{particle}} = -\frac{GM}{a} - \frac{GM}{a/2} = -\frac{3GM}{a}$. Magnitude is $\frac{3GM}{a}$.
A particle of mass $M$ is situated at the center of a spherical shell of same mass and radius $a$. The gravitational potential at a point situated at $\frac{a}{2}$ distance from the center, will be:
(2010 Pre)
The gravitational potential inside a shell is constant, $V_{\text{shell}} = -\frac{GM}{a}$. For particle, $V_{\text{particle}} = -\frac{GM}{a/2} = -\frac{2GM}{a}$. Total potential is $-\frac{3GM}{a}$.
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is
(2005)
Kinetic energy $K = \frac{GMm}{2r}$ and Potential energy $U = -\frac{GMm}{r}$. The ratio of their magnitudes is $|K|/|U| = \frac{1}{2}$.
A 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 $h = 3R$, $\Delta U = \frac{mg(3R)}{1+3} = \frac{3}{4}mgR$.
The 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 $v' = 4v$.
A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^6\text{ m}$ above the surface of earth. If earth’s radius is $6.38 \times 10^6\text{ m}$ and $g = 9.8\text{ m/s}^2$, then the orbital speed of the satellite is:
(2015 Re)
Orbital speed is calculated using $v = \sqrt{\frac{gR^2}{R+h}}$. Substituting the given values for $R$, $h$, and $g$, we get $v \approx 7.76\text{ km/s}$. Therefore, option B is correct.
A 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.
The radii of circular orbits of two satellites A and B of the earth, are $4R$ and $R$, respectively. If the speed of satellite A is $3V$, then the speed of satellite B will be:
(2010 Pre)
Orbital speed is inversely proportional to the square root of the radius ($v \propto \frac{1}{\sqrt{r}}$). Thus, $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}} = \sqrt{\frac{4R}{R}} = 2$, which means $v_B = 2 \times 3V = 6V$. Therefore, option C is correct.
For 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.