Question 61:
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 62:
moderateA 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 $$W_d = 200(1 - \frac{1}{2})$.n$W_d = 200 \times \frac{1}{2} = 100 \text{ N}$$.
Question 63:
moderateThe 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 $$ 1 + \frac{h}{R} = 4 \Rightarrow \frac{h}{R} = 3 \Rightarrow h = 3R$$.
Question 64:
moderateA 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$F = \frac{72}{2.25} = 32 \text{ N}$.
Question 65:
moderateA 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: $$E = \frac{3.0 \text{ N}}{0.06 \text{ kg}} = 50 \text{ N/kg}$$.
Question 66:
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 67:
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 $$\frac{gR^2}{r^2} = \omega^2r \Rightarrow r^3 = \frac{gR^2}{\omega^2} \Rightarrow r = (\frac{R^2g}{\omega^2})^{\frac{1}{3}}$$.
Question 68:
moderateA 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 $$W_h = \frac{72}{(1 + \frac{1}{2})^2} = \frac{72}{(\frac{3}{2})^2}$.n$W_h = 72 \times \frac{4}{9} = 32 \text{ N}$$.
Question 69:
moderateWhat 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 $d$: $$1 - \frac{d}{R} = \frac{1}{n} \Rightarrow \frac{d}{R} = \frac{n-1}{n} \Rightarrow d = \frac{R(n-1)}{n}$$.
Question 70:
moderateThe 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 $W = \frac{mgR}{1+1} = \frac{1}{2}mgR$.