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 $$W_d = 200(1 - \frac{1}{2})$.n$W_d = 200 \times \frac{1}{2} = 100 \text{ N}$$.
The acceleration due to gravity at a height $1 \text{ km}$ above the earth is the same as at a depth d below the surface of earth. Then:
(2017-Delhi)
For heights much smaller than the radius ($h \ll R$), $g_h \approx g(1 - \frac{2h}{R})$.nFor depth $d$, $g_d = g(1 - \frac{d}{R})$. Equating the two gives $1 - \frac{2h}{R} = 1 - \frac{d}{R}$.nThus $d = 2h$. Since $h = 1 \text{ km}$, $d = 2 \times 1 = 2 \text{ km}$.
The height at which the weight of a body becomes $1/16^{\text{th}}$, its weight on the surface of earth (radius R), is:
(2012 Pre)
Weight at height $h$ is given by $W_h = \frac{W}{(1 + \frac{h}{R})^2}$. Given $W_h = \frac{W}{16}$, we equate: $\frac{1}{16} = \frac{1}{(1 + \frac{h}{R})^2}$. Taking the square root gives $$ 1 + \frac{h}{R} = 4 \Rightarrow \frac{h}{R} = 3 \Rightarrow h = 3R$$.
A body of weight $72 \text{ N}$ moves from the surface of earth to a height half of the radius of the earth, then gravitational force exerted on it will be:
(2000)
Gravitational force (weight) at height $h$ is $F = \frac{W}{(1 + \frac{h}{R})^2}$.nSubstitute $h = \frac{R}{2}$ to get $F = \frac{72}{(1 + 0.5)^2}$.n$F = \frac{72}{2.25} = 32 \text{ N}$.
A body of mass $60 \text{ g}$ experiences a gravitational force of $3.0 \text{ N}$, when placed at a particular point. The magnitude of the gravitational field intensity at that point is:
(2022)
Gravitational field intensity $E$ is given by $E = \frac{F}{m}$.nConvert mass to kg: $m = 60 \text{ g} = 0.06 \text{ kg}$.nSubstitute the values: $$E = \frac{3.0 \text{ N}}{0.06 \text{ kg}} = 50 \text{ N/kg}$$.
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 $|K|/|U| = \frac{1}{2}$.