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 $v' = 4v$.
The 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 $W = \frac{mgR}{1+1} = \frac{1}{2}mgR$.
At what height from the surface of earth the gravitation potential and the value of $g$ are $-5.4 \times 10^{7} \text{ J kg}^{-1}$ and $6.0 \text{ ms}^{-2}$ respectively. Take the radius of earth as $6400 \text{ km}$:
(2016 – I)
Potential $V = -\frac{GM}{r} = -5.4 \times 10^{7}$ and $g = \frac{GM}{r^{2}} = 6.0$. Dividing $|V|$ by $g$ gives $r = 9000 \text{ km}$. Height $h = r - R = 9000 - 6400 = 2600 \text{ km}$.
Infinite number of bodies, each of mass $2 \text{ kg}$ are situated on x-axis at distances $1 \text{ m}$, $2 \text{ m}$, $4 \text{ m}$, $8 \text{ m}$, ….. respectively, from the origin. The resulting gravitational potential due to this system at the origin will be:
(2013)
Total potential $V = -GM \sum \frac{m}{r} = -2G (1 + \frac{1}{2} + \frac{1}{4} + ...) = -2G \times \frac{1}{1 - 0.5} = -4G$.
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 satellite of mass $m$ is orbiting the earth (of radius $R$) at a height $h$ from its surface. The total energy of the satellite in terms of $g_0$, the value of acceleration due to gravity at the earth’s surface, is:
(2016 – II)
Total energy of a satellite is the sum of kinetic and potential energy, given by $E = -\frac{GMm}{2(R+h)}$. Substituting $GM = g_0 R^2$, we obtain $E = -\frac{mg_0 R^2}{2(R+h)}$. Thus, option D is correct.
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.