Physics MCQs for NEET — Practice Questions with Answers

The weight of an astronaut in an artificial satellite revolving around the Earth is zero. This is because the astronaut and the satellite are both in free fall towards the Earth, creating a condition of apparent weightlessness. The gravitational force provides the centripetal force needed to keep the satellite in orbit, but since the astronaut is also in free fall, they do not experience this force as weight.

6R from the surface of earth and 7R from the center

Gravitational force provides the required centripetal force for orbiting the satellite $ { m \upsilon^2 \over R } = { K \over R } $ because $ F \alpha { 1 \over R } $ $ \therefore \upsilon \alpha R^o $

Total energy = -K.E. = $ {-1/2 m \upsilon^2 } $

B.E. = -K.E. And it this amount of energy(Ek) given to satellite it will escape into outer space

$ P.E = { GMm \over r } = - { GMm \over Re+h } = -{ GMm \over 2 Re } = - {g Re^2 m \over 2 Re } = - { 1 \over 2 } mg Re = 0.5 mg Re $

$ B.E = { - GMm \over r } $ if B.E. decreases the r also decreases and V increases as $ v \alpha { 1 \over r } $

$ {V_B \over V_A } = \sqrt { \gamma_A \over \gamma_B } = \sqrt { 4R \over R } = 2 $

For satellites in circular orbits, the orbital speed is inversely proportional to the square root of the orbital radius. If satellite A has a speed of 3v at a radius of 4R, then the speed of satellite B, which has a radius R, can be calculated as follows:

$$v_B = v_A imes rac{ ext{sqrt}(R_A)}{ ext{sqrt}(R_B)} = 3v imes rac{ ext{sqrt}(4R)}{ ext{sqrt}(R)} = 3v imes 2 = 6v.$$

Therefore, the speed of satellite B is 6v.

time period of revolution of moon around the earth = 1 lunar month $ {T_e \over T_m } = \left( {r_e \over r_m } \right)^{3/2} = {1 \over 2 } ^ {3/2} = 2^{-2/3} $