## A particle of mass m is placed at the centre of a uniform spherical shell of equal mass and radius a…

Q: A particle of mass m is placed at the centre of a uniform spherical shell of equal mass and radius a. Find the gravitational potential at a point P at a distance a/2 from the centre.

Sol: The gravitational potential at P due to particle at centre is

$\large V_1 = \frac{-Gm}{a/2} = – \frac{2 G m}{a}$

The potential at P due to shell is

$\large V_2 = – \frac{ G m}{a}$

The net potential at P is $\large V = V_1 + V_2 = – \frac{3 G m}{a}$

## The gravitational field due to a mass distribution is given by E = – K/x3 in X- direction…

Q: The gravitational field due to a mass distribution is given by E = – K/x3 in X- direction. Taking the gravitational potential to be zero at infinity, find its value at a distance x.

Sol: The potential at a distance x is

$\large V = -\int E dx$

$\large V = \int_{\infty}^{x} \frac{K}{x^3} dx$

$\large V = [-\frac{K}{2x^2}]_{\infty}^{x}$

$\large V = \frac{-K}{2 x^2}$

## Two bodies of masses 100 kg and 10,000 kg are at a distance of 1 m apart. At what distance from 100 kg on the line joining them…

Q: Two bodies of masses 100 kg and 10,000 kg are at a distance of 1 m apart. At what distance from 100 kg on the line joining them will the resultant gravitational field intensity be zero ?

Sol: $\large \frac{G \times 100}{x^2} = \frac{G \times 10000}{(1-x)^2}$

$\large \frac{1}{x} = \frac{10}{1-x}$

10 x = 1-x

$\large x = \frac{1}{11} m$

## A man can jump 1.5 m on the Earth. Calculate the approximate height he might be able to jump on a planet whose density…

Q: A man can jump 1.5 m on the Earth. Calculate the approximate height he might be able to jump on a planet whose density is one-quarter that of the Earth and whose radius is one-third that of the Earth.

Sol: We know that, in case of Earth,

$\large g = \frac{G M}{R^2}$

$\large g = \frac{G}{R^2}\frac{4}{3}\pi R^3 \rho$

$\large g = (\frac{4}{3}\pi G ) R \rho$

Similarly, for the other planet whose radius R/3 and density ρ/4

$\large g’ = (\frac{4}{3}\pi G ) (R/3) (\rho/4 )$

$\large g’ = \frac{g}{12}$

$\large h_{max} = \frac{u^2}{2 g}$

$\large \frac{h’}{h} = \frac{g}{g’} = 12$

h’ = 12 h

h’ = 12 × 1.5 = 18 m

## Find the percentage decrease in the weight of the body when taken to a depth of 32 Km below the surface of earth.

Q: Find the percentage decrease in the weight of the body when taken to a depth of 32 Km below the surface of earth.

Sol: Weight of the body at depth d is

$\large m g’ = m g(1-\frac{d}{R})$

% decrease in weight $\large = \frac{mg-mg’}{mg}\times 100$

$\large = \frac{d}{R}\times 100$

$\large = \frac{32}{6400}\times 100$

= 0.5 %

## How much above the surface of earth does the acceleration due to gravity reduce by 36% of its value on the surface of earth…

Q: How much above the surface of earth does the acceleration due to gravity reduce by 36% of its value on the surface of earth.

Sol: Since g reduces by 36%, the value of g there is 100-36 = 64%.

$\large g’ = \frac{64}{100}g$

If h is the height of location above the surface of earth , then

$\large g’ = \frac{g R^2}{(R+h)^2}$

$\large \frac{64}{100}g = \frac{g R^2}{(R+h)^2}$

$\large \frac{8}{10} = \frac{R}{R+h}$

$\large h = \frac{R}{4} = \frac{6.4 \times 10^6}{4}$

= 1.6 × 106 m

## The height at which the acceleration due to gravity becomes g/9 in terms of the radius of the earth (R) is

Q: The height at which the acceleration due to gravity becomes g/9 (where g is the acceleration due to gravity on the surface of the earth) in terms of the radius of the earth (R) is

Sol: $\large g’ = g(\frac{R}{R+h})^2$

$\large \frac{g}{9} = g(\frac{R}{R+h})^2$

$\large \frac{1}{3} = \frac{R}{R+h}$

3R = R+h

⇒ 2R = h.

## What is the time period of rotation of the earth around its axis so that objects at the equator becomes weightless ?

Q: What is the time period of rotation of the earth around its axis so that objects at the equator becomes weightless ? (g=9.8 m⁄s2,Radius of earth = 6400 km))

Sol: When earth is rotating the apparent weight of a body at the equator is given by

Wapp = mg – m R ω2

If bodies are weightless at the equator

0 = mg – m R ω2

g = R ω2

$\large \omega = \sqrt{\frac{g}{R}}$

Time period, $\large T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g}}$

$\large = 2\pi \sqrt{\frac{6.4 \times 10^6}{9.8}}$

= 5078 sec = 84 min 38 sec

## A star 2.5 times the mass of the sun is reduced to a size of 12 km and rotates with a speed of 1.5 rps…

Q: A star 2.5 times the mass of the sun is reduced to a size of 12 km and rotates with a speed of 1.5 rps. Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the Sun = 2 × 1030 kg).

Sol: Acceleration due to gravity, g = GM/R2

$\large g = \frac{6.67 \times 10^{-11} \times 2.5 \times 2 \times 10^{30}}{(12000)^2}$

= 2.3 × 1012 m/s2

Centrifugal acceleration

$\large = r \omega^2 = r(2 \pi f)^2$

$\large = 12000(2 \pi \times 1.5)^2$

= 1.1 × 106 m/s2

Since, g > r ω 2,the body will remain stuck with the surface of star.

## The mean orbital radius of the Earth around the Sun is 1.5×108 km. Estimate the mass of the Sun.

Q: The mean orbital radius of the Earth around the Sun is 1.5 × 108 km. Estimate the mass of the Sun.

Sol: As the centripetal forces is provided by the gravitational pull of the Sun on the Earth

$\large \frac{G M_s M_e}{r^2} = M_e r \omega^2$

$\large \frac{G M_s M_e}{r^2} = M_e r (\frac{2\pi}{T})^2$

$\large M_s = \frac{4 \pi^2 r^3}{G T^2}$

Given, r = 1.5 × 108 Km = 1.5 × 1011 m ;

T = 365 days = 365 × 24 × 60 × 60 s

$\large M_s = \frac{4 (22/7)^2 \times (1.5 \times 10^{11})^3}{6.67 \times 10^{-11}\times (365 \times 24 \times 60 \times 60)^2}$

≈ 2 × 1030 kg