Gravitation

Q: The mass and diameter of a planet are twice those of earth . What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on the earth ?

(a) $\sqrt{2} sec$

(b) $2 \sqrt{2} sec$

(c) $\frac{1}{\sqrt{2}} sec$

(d) $\frac{1}{2 \sqrt{2}} sec$

Solution : T_e = 2 sec (Having seconds pendulum)

$\displaystyle T = 2 \pi \sqrt{\frac{l}{g}}$

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

$\displaystyle T = 2 \pi \sqrt{\frac{l}{\frac{G M}{R^2} }}$

$\displaystyle T = 2 \pi \sqrt{\frac{l R^2}{G M}}$

$\displaystyle \frac{T_p}{T_e} = \sqrt{\frac{R_p^2}{R_e^2} \times \frac{M_e}{M_p}}$

$\displaystyle \frac{T_p}{T_e} = \sqrt{\frac{4 R_e^2}{R_e^2} \times \frac{M_e}{2 M_e}}$

$\displaystyle \frac{T_p}{T_e} = \sqrt{2} $

$\displaystyle T_p = \sqrt{2} T_e$

$\displaystyle T_p = \sqrt{2} \times 2$

$\displaystyle T_p = 2 \sqrt{2} sec $

Correct option is (b)

Q: A satellite is launched into a circular orbit of radius R round the earth. A second satellite is launched into an orbit of radius (1.01)R . The period of the second satellite is larger than that of the first one by approximately

(a) 0.5 %

(b) 1.0 %

(c) 1.5 %

(d) 3.0 %

Ans: (c)

Sol: $T^2 \propto R^3 $

$(\frac{T’}{T})^2 = (\frac{1.01 R}{R})^3 $

$(\frac{T’}{T}) = (\frac{1.01 R}{R})^{3/2} $

$\frac{T’}{T} = (1 + \frac{1}{100})^{3/2} $

$\frac{T’}{T} = (1 + \frac{3}{2}\frac{1}{100}) $ (Using Binomial Theorem )

$\frac{T’}{T} -1 = (\frac{3}{2}\frac{1}{100}) $

$\frac{T’ – T}{T} = (\frac{3}{2} \times \frac{1}{100}) $

$(\frac{T’-T}{T} )\times 100 = (\frac{3}{2}\times \frac{1}{100}) \times 100 $

= 1.5 %

Q: If r denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to

(a) r^3/2

(b) r

(c) √r

(d) r²

Ans: (c)

Sol: Angular Momentum , L = m v r

$L = m \sqrt{\frac{G M}{r}} r $

$L \propto \sqrt{r} $

Q: How long will a satellite, placed in a circular orbit of radius that is (1/4)^th the radius of a geostationary satellite, take to complete one revolution around the earth

(a) 12 hours

(b) 6 hours

(c) 3 hours

(d) 4 days

Ans: (c)

Sol: As , $T^2 \propto r^3 $

$ (\frac{T’}{T})^2 = (\frac{r’}{r})^3 = (\frac{r/4}{r})^3 $

$ (\frac{T’}{T})^2 = \frac{1}{64} = \frac{}{}$

$ (\frac{T’}{T}) = \frac{1}{8} $

T’ = 3 hours

Q: In planetary motion the areal velocity of position vector of a planet depends on angular velocity (ω) and the distance of the planet from sun (r). If so the correct relation for areal velocity is

(a) $\frac{dA}{dt} \propto \omega r$

(b) $\frac{dA}{dt} \propto \omega^2 r$

(c) $\frac{dA}{dt} \propto \omega r^2 $

(d) $\frac{dA}{dt} \propto \sqrt{\omega r} $

Ans: (c)

Sol: $\frac{dA}{dt} = \frac{d}{dt} (\frac{1}{2}) r (r \theta ) $

$\frac{dA}{dt} = (\frac{1}{2}) r^2 (\frac{d \theta }{dt} ) $

$\frac{dA}{dt} = (\frac{1}{2}) r^2 \omega $

Q: A satellite close to earth is in orbit above the equator with a period of rotation of 1.5 hrs. If it is above a point P on the equator at some time, it will be above P again after time

(A) 1.5 hrs.

(B) 1.6 hrs. if it is rotating from west to east.

(C) 24/17 hrs. if it is rotating from west to east.

(D) 24/17 hrs. if it is rotating from east to west.

Solution : Let ω_o = the angular velocity of earth above its axis = 2π/24  rad/hr

Let ω = angular velocity of satellite.

ω = 2π/1.5

For a satellite rotating from west to east. (same as earth), the relative angular velocity ω₁ = ω – ω_o

Time period of rotation relative to earth = 2π/ωω₁ = 1.6 h

Now for a satellite rotating from east to west (opposite to earth) the relative angular velocity ω₂ = ω + ω_o

Time period of rotation relative to earth =2π/ω₂ = 24/17 hrs.

Q: The earth revolve round the sun in one year. If the distance between them becomes double, the period of revolution will be

(a) 1/2 year

(b) 2√2 years

(c) 4 years

(d) 8 years

Ans: (b)

Sol: According to Kepler’s 3rd Law (Law of Time Period)

$\displaystyle (\frac{T_2}{T_1})^2 = (\frac{r_2}{r_1})^3$

$\displaystyle (\frac{T_2}{T_1})^2 = (\frac{2 r_1}{r_1})^3$

$\displaystyle (\frac{T_2}{T_1})^2 = 8 $

$\displaystyle (\frac{T_2}{T_1}) = 2 \sqrt{2} $

$\displaystyle T_2 = 2 \sqrt{2} T_1 $

= 2√2 years

Problem :  There are two fixed heavy masses of magnitude M of high density on x-axis at a distance 2d apart.  A small mass m moves in a circle of radius R in the y-z plane between the heavy masses.  Find the speed of the small particle.

Solution :

Force of attraction between M and m is

$ \displaystyle F = \frac{G M m}{R^2 + d^2} $

By symmetry F_x components will cancel.

  ∴ The net force, which provides the centripetal force, is given by

$ \displaystyle 2F_y = 2.\frac{G M m}{R^2 + d^2} \frac{R}{\sqrt{R^2 + d^2}} $

$ \displaystyle = \frac{2 G M m R}{(R^2 + d^2)^{3/2}} $

$ \displaystyle \frac{2 G M m R}{(R^2 + d^2)^{3/2}} = \frac{m v^2}{R}$

$ \displaystyle v = \sqrt{\frac{2 G M R^2}{(R^2 + d^2)^{3/2}}} $

Problem :  Two satellites of same mass are launched in the same orbit around the earth so that they rotate opposite to each other.  If they collide in-elastically, obtain the total energy of the system before and just after the collision.  Describe the subsequent motion of the wreckage .

Solution:

 Potential energy of the satellite in its orbit is

$ \displaystyle U = -\frac{G M m}{r}$

$ \displaystyle K.E = |\frac{U}{2}| = \frac{G M m}{2 r} $

where m is mass of satellite, M  the mass of the earth and r the orbital radius.

Total energy of one satellite = K. E. + P.E.

$ \displaystyle E = – \frac{G M m}{2 r} $

For two satellites, total energy $ \displaystyle E = – \frac{G M m}{ r} $

Let v’ be the velocity after collision. By conservation of momentum

$ \displaystyle m\vec{v_1} + m \vec{v_2} = 0 = (m+m)\vec{v’} $

⇒      v’ = 0

The wreckage of mass (2m) has no kinetic energy, but it has only potential energy.  So, energy after collision

$ \displaystyle = – \frac{G M (2m)}{ r} $

The wreckage falls down under gravity.

Problem  :  A uniform sphere has a mass M and radius R.  Find the gravitational  pressure P inside the sphere, as a function of the distance r from its centre.

Solution :

Consider a layer of thickness dr at a distance r from the centre of the sphere.
Now force due to the layer dF = 4πr²dP

$ \displaystyle dp (4\pi r^2) = \frac{G (\frac{4}{3})\pi r^3 \rho (4 \pi r^2 dr \rho)}{r^2} $

(Where ρ is the mean density of sphere)

$ \displaystyle dp = G \frac{4}{3}\pi \rho^2 r dr $

$ \displaystyle P = G \int_{r}^{R} \frac{4}{3}\pi \rho^2 r dr $

$ \displaystyle P = G \frac{2 \pi}{3}\rho^2 (R^2 – r^2) $

$ \displaystyle = \frac{3}{8}G \frac{(1- r^2/R^2)M^2}{\pi R^4} $

[ As, $ \displaystyle \rho = \frac{M}{\frac{4}{3}\pi R^3} $]