Question 11:
moderateThe radius of earth is about $6400\text{ km}$ and that of planet mars is $3200\text{ km}$. The mass of the earth is about $10$ times mass of planet mars. An object weighs $200\text{ N}$ on the surface of earth. Its weight on the surface of planet mars will be:
(1994)
Gravity $g \propto \frac{M}{R^2}$. The ratio of weights is $W_m/W_e = (M_m/M_e) \times (R_e/R_m)^2 = (1/10) \times (6400/3200)^2 = 0.4$. Thus, the weight on Mars is $W_m = 0.4 \times 200 = 80\text{ N}$.
Question 12:
easyA 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)
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 $g = \frac{GM_P}{(D_P/2)^2} = \frac{4GM_P}{D_P^2}$.
Question 13:
easyImagine 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)
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 $g' = 3g$.
Question 14:
easyThe 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)
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 $R_P = \frac{1}{2}R$.
Question 15:
easyTwo spheres of masses $m$ and $M$ are situated in air and the gravitational force between them is $F$. The space around the masses is now filled with a liquid of specific density $c$. The gravitational force will now be:
(2003)
The gravitational force between two point masses is completely independent of the intervening medium. Therefore, the force remains $F$ even when the space is filled with a liquid.
Question 16:
difficultThe 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)
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 $9 g_B \times 2 = g_B \times h_B \implies h_B = 18\text{ m}$.
Question 17:
moderateFor moon, its mass is $\frac{1}{81}$ of earth mass and its diameter is $\frac{1}{3.7}$ of earth diameter. If acceleration due to gravity at earth surface is $9.8\text{ m/s}^2$ then at moon its value is:
(1999)
Using $g \propto \frac{M}{R^2}$, we have $g_m = g_e \times (\frac{M_m}{M_e}) \times (\frac{R_e}{R_m})^2$. Substituting the values: $g_m = 9.8 \times \frac{1}{81} \times (3.7)^2 \approx 1.65\text{ m/s}^2$.
Question 18:
easyThe 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)
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 $\rho = \frac{3g}{4\pi GR}$.
Question 19:
moderateIf 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)
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 $v \propto R^0$.
Question 20:
moderateThe 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)
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 $$\frac{gR^2}{r^2} = \omega^2r \Rightarrow r^3 = \frac{gR^2}{\omega^2} \Rightarrow r = (\frac{R^2g}{\omega^2})^{\frac{1}{3}}$$.