Question 191:
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 192:
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 193:
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 194:
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 195:
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 196:
difficultTwo particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed $v$ of each particle is:
(1995)
The gravitational force provides the necessary centripetal force. $\frac{mv^2}{R} = \frac{Gmm}{(2R)^2} = \frac{Gm^2}{4R^2}$. Solving for $v$, we get $v^2 = \frac{Gm}{4R}$, which means $v = \frac{1}{2}\sqrt{\frac{Gm}{R}}$.
Question 197:
moderateThe earth (mass $= 6 \times 10^{24}\text{ kg}$) revolves around the sun with an angular velocity of $2 \times 10^{-7}\text{ rad/s}$ in a circular orbit of radius $1.5 \times 10^8\text{ km}$. The force exerted by the sun on the earth, in newton, is:
(1995)
The force is the centripetal force $F = mR\omega^2$. Substituting the values: $F = (6 \times 10^{24}) \times (1.5 \times 10^{11}\text{ m}) \times (2 \times 10^{-7})^2 = 9 \times 10^{35} \times 4 \times 10^{-14} = 36 \times 10^{21}\text{ N}$.
Question 198:
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 199:
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 200:
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 $$W_d = 200(1 - \frac{1}{2})$.n$W_d = 200 \times \frac{1}{2} = 100 \text{ N}$$.