A spherical planet has a mass $M_P$ and diameter $D_P$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to:
(2012 Pre)
1. $\frac{4GM_P}{D_P^2}$
2. $\frac{GM_Pm}{D_P^2}$
3. $\frac{GM_P}{D_P^2}$
4. $\frac{4GM_Pm}{D_P^2}$
View Answer
Acceleration due to gravity is given by $g = \frac{GM_P}{R_P^2}$. Substituting the radius as half of the diameter, $R_P = \frac{D_P}{2}$, we get $g = \frac{GM_P}{(D_P/2)^2} = \frac{4GM_P}{D_P^2}$.
Imagine a new planet having the same density as that of earth but it is $3$ times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is $g$ and that on the surface of the new planet is $g’$, then:
(2005)
1. $g' = 3g$
2. $g' = \frac{g}{9}$
3. $g' = 9g$
4. $g' = \frac{g}{3}$
View Answer
Acceleration due to gravity in terms of density is $g = \frac{4}{3}\pi \rho G R$. Since density $\rho$ is constant, $g \propto R$. For a planet $3$ times bigger in size ($R' = 3R$), the new gravity is $g' = 3g$.
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$, the radius of the planet would be:
(2004)
1. $4R$
2. $\frac{1}{4}R$
3. $\frac{1}{2}R$
4. $2R$
View Answer
Using $g = \frac{4}{3}\pi G \rho R$, we have $\rho_P R_P = \rho_E R_E$ since $g$ is the same for both. Given $\rho_P = 2\rho_E$, we get $2\rho_E R_P = \rho_E R$, which gives $R_P = \frac{1}{2}R$.
The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B$. A man jumps to a height of $2\text{ m}$ on the surface of $A$. What is the height of jump by the same person on the planet $B$:
(2003)
1. $\frac{2}{9}\text{ m}$
2. $18\text{ m}$
3. $6\text{ m}$
4. $\frac{2}{3}\text{ m}$
View Answer
The muscular work done in jumping is the same, so the potential energy gained is constant: $m g_A h_A = m g_B h_B$. Substituting $g_A = 9 g_B$ and $h_A = 2\text{ m}$, we get $9 g_B \times 2 = g_B \times h_B \implies h_B = 18\text{ m}$.
The acceleration due to gravity $g$ and mean density of the earth $\rho$ are related by which of the following relations? (where $G$ is the gravitational constant and $R$ is the radius of the earth.):
(1995)
1. $\rho = \frac{3g}{4\pi GR}$
2. $\rho = \frac{3g}{4\pi GR^3}$
3. $\rho = \frac{4\pi gR^2}{3G}$
4. $\rho = \frac{4\pi gR^3}{3G}$
View Answer
We know $g = \frac{GM}{R^2}$ and $M = \frac{4}{3}\pi R^3 \rho$. Substituting $M$, we get $g = \frac{G}{R^2} \times \frac{4}{3}\pi R^3 \rho = \frac{4}{3}\pi G \rho R$. Rearranging for density gives $\rho = \frac{3g}{4\pi GR}$.
If the gravitational force between two objects were proportional to $\frac{1}{R}$ (and not as $\frac{1}{R^2}$), where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to:
(1994, 89)
1. $R$
2. $R^0$ (independent of R)
3. $\frac{1}{R^2}$
4. $\frac{1}{R}$
View Answer
Centripetal force is provided by the given gravitational force: $$\frac{mv^2}{R} = \frac{k}{R}$$.
Solving for $v$, we get $v^2 = \frac{k}{m}$.nSince $k$ and $m$ are constants, $v$ is independent of $R$, meaning $v \propto R^0$.
The mean radius of earth is R, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at earth’s surface is g. What will be the radius of the orbit of a geostationary satellite?
(1992)
1. $(\frac{R^2g}{\omega^2})^{\frac{1}{3}}$
2. $(\frac{Rg}{\omega^2})^{\frac{1}{3}}$
3. $(\frac{R^2\omega^2}{g})^{\frac{1}{3}}$
4. $(\frac{R^2g}{\omega})^{\frac{1}{3}}$
View Answer
Gravitational force provides the centripetal force: $$\frac{GMm}{r^2} = m\omega^2r$$.nSince $$g = \frac{GM}{R^2}$$, we can write $$GM = gR^2$$.nSubstituting this gives $$\frac{gR^2}{r^2} = \omega^2r \Rightarrow r^3 = \frac{gR^2}{\omega^2} \Rightarrow r = (\frac{R^2g}{\omega^2})^{\frac{1}{3}}$$.