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)
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)
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)
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)
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 $9 g_B \times 2 = g_B \times h_B \implies h_B = 18\text{ m}$.
For 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: $g_m = 9.8 \times \frac{1}{81} \times (3.7)^2 \approx 1.65\text{ m/s}^2$.
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)
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 $\rho = \frac{3g}{4\pi GR}$.
The 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 $W_m = 0.4 \times 200 = 80\text{ N}$.
A body weighs $72 \text{ N}$ on the surface of the earth. What is the gravitation force on it, at a height equal to half the radius of the earth?
(2020)
The weight at height $h$ is given by $$W_h = \frac{W}{(1 + \frac{h}{R})^2}$$.nSubstituting $h = \frac{R}{2}$, we get $$W_h = \frac{72}{(1 + \frac{1}{2})^2} = \frac{72}{(\frac{3}{2})^2}$.n$W_h = 72 \times \frac{4}{9} = 32 \text{ N}$$.
What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth = R)
(2020-Covid)
The acceleration due to gravity at depth $d$ is $g_d = g(1 - \frac{d}{R})$.nGiven $g_d = \frac{g}{n}$, we have $\frac{g}{n} = g(1 - \frac{d}{R})$. Solving for $d$: $$1 - \frac{d}{R} = \frac{1}{n} \Rightarrow \frac{d}{R} = \frac{n-1}{n} \Rightarrow d = \frac{R(n-1)}{n}$$.
A body weighs $200 \text{ N}$ on the surface of the earth. How much will it weigh half way down to the centre of the earth?
(2019)
The weight at depth $d$ is $W_d = W(1 - \frac{d}{R})$.nGiven $d = \frac{R}{2}$ (halfway to the center), we have $$W_d = 200(1 - \frac{1}{2})$.n$W_d = 200 \times \frac{1}{2} = 100 \text{ N}$$.