Gravitation Test - 14

For central force, torque is zero.

\(\mathrm{\therefore \tau=\frac{dL}{dt}=0\Rightarrow L=constant}\)

i.e. Angular momentum is constant.

\(T^2 \propto R^3\)

\(\therefore (\frac{T_1}{T_2})^2 = (\frac{R_1}{R_2})^3\)

\(\therefore T_2 \sqrt{(\frac{R_2}{R_1})^3.T_1^2}\)

= \(\sqrt{(\frac{2R}{R})^3.1}\)

 \(\because T_1 = 1 year\)

= \(\sqrt 8\)

= 2.8 years

Given,

\(T_A = 8T_B\)

\((\frac{T_A}{T_B})^2 = (\frac{R_A}{R_B})^3\)

\(\frac{R_A}{R_B} = (\frac{T_A}{T_B})^{\frac{2}{3}}\)

\(\frac{R_A}{R_B} = (\frac{8T_B}{T_B})^{\frac{2}{3}}\)

\(\frac{R_A}{R_B} = 2 ^{3 \times \frac{2}{3}}\) = \(2^2 = 4\)

\(R_A = 4R_B\)

\(T^2 \propto r^3\)

Since universal gravitation constant, G. is decreasing with time, gravitational force would decrease considering the fact that conservation of energy and momentum of earth-satellite system is still valid, decrease in gravitational force would cause the separation between the components of system to increase and angular velocity to reduce, product \(\mathrm{'\omega r'}\) remaining constant.

Since tangential velocity, \(\mathrm{v = \omega r}\), therefore 'V' is the quantity that remains unchanged.

(i) Comet does not have elliptical orbit. Hence doesn’t obey Kepler's law of planetary action.

(ii) They move along paths which are hyperbolic or almost parabolic.

Escape velocity is giveb by formula

\(v_e = \sqrt{\frac{2GM}{R}}\)

Now gravitational potential energy

\(U_G = {\frac{-GMm}{R}}\)

\(U_G = -\frac{1}{2}mv_e^2\)

\(U_G = -\frac{1 \times 100^2}{2}\)

\(U_G= - 5000J\)

\(h = \frac{R}{\frac{V_e^2}{V^2} - 1}\)

= \(\frac{R}{(\frac{11.2}{10})^2 -1} \dots V_e = 11.2 km/s\)

= h = 3.93R

h ≈ 4R

acceleration due to gravity is g = \(\frac{GM}{R^2}\)    [R = radius of earth,

G = gravitational constant, M = mass of earth]

 if radius shrinks half = \(\frac{R}{2}\)

\(g = \frac{GM}{(\frac{R}{2})^2}\)

\(g = 4 \times \frac{GM}{R^2}\)

Acceleration due to Gravity increases 4 times.

Inertial mass is free from gravitational force. It depends upon only mass. Gravitational mass is dependent on gravitational force.

The Cavendish experiment, performed in 1797-1798 by British scientist Henry Cavendish, was the first experiment to measure the force of gravity between masses in the laboratory and the first to yield accurate values for the gravitational constant.

Acceleration of anything falling freely under gravity does not depend upon their masses or type of material.

so both have same acceleration and hence take same time to reach the ground.

As, The earth is not a perfect sphere, The gravity keeps on changing as we move from one place to another.

But it is maximum at the place where it is nearer to the centre. so, Earth’s gravity is the maximum at the poles because the Earth is kind of an ellipse (not a perfect sphere.)

And the equator is further away from the centre of mass of the Earth than at the poles.

so, As we go from equator to poles, the value of g Increases.

g' = \(g(\frac{R}{(R+H})^2 \)  .......at height H, acceleration = g'

g' = \(g \Big(\frac{R}{R + (\frac{R}{2})}\Big)^2\)

g' \(= g(\frac{4}{9})\)

W = mg' = \(\frac{4}{9}\) mg 

Weight on earth

= \(\frac{4}{9}\) x 72

= 32N

For two-electron \(\frac{F_g}{F_e} = 10^{-43}\) i.e. gravitational force is negligible in comparison to electrostatic force of attraction.

Kinetic and potential energies vary with position of earth w.r.t. sun. Angular momentum remains constant everywhere.

As observed from the earth, the sun appears to move in an approximately circular orbit. The gravitational force of attraction between the earth and the sun always follows inverse square law. Due to relative motion between the earth and mercury, the orbit of mercury, as observed from the earth, will not be approximately circular, since the major gravitational force on mercury is due to the sun.

\(V_e = \sqrt2 v_0\) i.e. if the orbital velocity of moon is increased by factor of \(\sqrt 2\) then it will escape out from the gravitational field of earth.

\(g = \frac{GM}{R^2}\)

Given (\(M_e = 81 M_m, R_e = 3.5 R_m\)) Substituting the above values, \(\frac{g_m}{g_e} \) = 0.15

Archimedes uplift as it depends on the weight of the body.