Q: Consider a star of mass m_2 ~kg revolving in a circular orbit around another star of mass m_1 ~kg with m_1 m_2. The heavier star slowly acquires mass from the lighter star at a constant rate of kg / s

$ aligned & v=v_0= G m_1 r \\ & d v_0 d t = v_0 2 [ m_1 - 1 r d r d t ] aligned $ $ aligned & F =m_2 d v d t \\ & G m_1 r^2 v_r v = d v d t \\ & v_0 r d r d t 1 v_0 = d v d t aligned $ $ V_0 d d t = X_0 2 [ r m_1 - 1 r d r d t ]$ $ 3 2 1 d d t = r 2 m_1 1 d r d t = 3 m_2 $

Q: A particle of mass m is under the influence of the gravitational field of a body of mass M( m). The particle is moving in a circular orbit of radius r_0 with time period T_0 around the mass M. Then, t

aligned & F _1= GMm r _0^2 \\ & ~F _2= GMm r _0^2 - 3 ~m r _0^4 \\ & _1^2 _0^2 = F _2 ~F _1 = GM r _0^2 - 3 r _0^4 GM r _0^2 \\ & T _0^2 ~T _1^2 =1- 3 GMr _0^2 \\ & T _1^2- T _0^2 ~T _1^2 = 3 GMr _0^2 aligned

Q: The distance between two stars of masses 3 M S and 6 M S is 9 R . Here R is the mean distance between the centres of the Earth and the Sun, and M S &nbs

T = 2 π R 3 / 2 G M S For binary stars, T ' = 2 π r 3 / 2 G M 1 + M 2 Put M 1 = 3 M S , M 2 = 6 M S and r = 9 R T ' = 2 π × 9 3 / 2 × R 3 / 2 G 3 M S + 6 M S = 2 π × 9 3 / 2 × R 3 / 2 9 1 / 2 G M S = 9 2 π R 3 / 2 G M S ∴ T ' = 9 T Value of n = 9 .

Question 1

A particle of mass m, and angular momentum is moving in a circular orbit of radius r_0 under the influence of an attractive force F (r) = - k r^2 r . Keeping its angular momentum unchanged, the partic

Accepted Answer

The equation of motion for the radial distance r is given by: m d^2r dt^2 = F(r) + ^2 mr^3 = - k r^2 + ^2 mr^3 For a circular orbit of radius r_0, the radial acceleration is zero: - k r_0^2 + ^2 mr_0^3 = 0 r_0 = ^2 mk Let the particle be displaced by a small distance x such that r = r_0 + x. The restoring force is: m d^2x dt^2 = - k (r_0+x)^2 + ^2 m(r_0+x)^3 Using the binomial expansion for x r_0: m d^2x dt^2 - k r_0^2 (1 - 2x r_0 ) + ^2 mr_0^3 (1 - 3x r_0 ) Since k r_0^2 = ^2 mr_0^3 , the const

Question 2

Consider a star of mass m_2 ~kg revolving in a circular orbit around another star of mass m_1 ~kg with m_1 m_2. The heavier star slowly acquires mass from the lighter star at a constant rate of kg / s

Accepted Answer

$ aligned & v=v_0= G m_1 r \ & d v_0 d t = v_0 2 [ m_1 - 1 r d r d t ] aligned $ $ aligned & F =m_2 d v d t \ & G m_1 r^2 v_r v = d v d t \ & v_0 r d r d t 1 v_0 = d v d t aligned $ $ V_0 d d t = X_0 2 [ r m_1 - 1 r d r d t ]$ $ 3 2 1 d d t = r 2 m_1 1 d r d t = 3 m_2 $

Question 3

A geostationary satellite above the equator is orbiting around the earth at a fixed distance r_1 from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite di

Accepted Answer

T r ^ 3 / 2 T _2 ~T _1 = ( r _2 r _1 )^ 3 / 2 _2 _1 = ( r_1 r_2 )^ 3 / 2 =(1.21)^ 3 / 2 aligned & _2= _1(1.331) ...(i)\ & ( _2+ _1 ) t_0=2 ...(ii) aligned t_0= 2 _2+ _1 = 2 ( 4 3 +1 ) _1 = 6 7 _1 t _0= 6 2 ( ~T _ GSS ) (7) t _0= 3 24 hours 7 = 24 p hours p = 7 3 =2.33

Question 4

A particle of mass m is under the influence of the gravitational field of a body of mass M( m). The particle is moving in a circular orbit of radius r_0 with time period T_0 around the mass M. Then, t

Accepted Answer

aligned & F _1= GMm r _0^2 \ & ~F _2= GMm r _0^2 - 3 ~m r _0^4 \ & _1^2 _0^2 = F _2 ~F _1 = GM r _0^2 - 3 r _0^4 GM r _0^2 \ & T _0^2 ~T _1^2 =1- 3 GMr _0^2 \ & T _1^2- T _0^2 ~T _1^2 = 3 GMr _0^2 aligned

Question 5

Two satellites P and Q are moving in different circular orbits around the Earth(radius R ). The heights of P and Q  from the Earth surface are h P  and h Q , respectively, where h P = R 3 .

Accepted Answer

Acceleration due to gravity at any height from the Earth's surface is given by,  g = G M r 2 , where r = distance from the centre of the Earth. Now, we can write g P g Q = G M r P 2 G M r Q 2 &#8658; 36 25 = r Q r P 2 &#8658; r Q r P = 6 5 &#8658; r Q = 6 5 &#215; R + R 3 &#8658; R + h Q = 8 5 R &#8658; h Q = 3 5 R

Question 6

Two spherical stars A  and B  have densities &#961; A and &#961; B , respectively. A  and B  have the same radius, and their masses M A and M B are related by M B = 2 M A . Due to

Accepted Answer

Given here:  R A = R B = R  and  M B = 2 M A . Now, after interaction process, radius of remaining star  A  is  R A ' = R 2  and its mass is  M A ' = &#961; A 4 3 &#960; R A 2 3 = M A 8 Applying conservation of energy,   - G M A ' m R A ' + 1 2 m v A 2 = 0 Escape velocity of star  A  is  v A = 2 G M A 8 &#215; R 2 = v 0 2 Now, for  B ,  mass collected over  B  is  M B ' = M A - M A 8 = 7

Question 7

The distance between two stars of masses  3 M S  and  6 M S  is  9 R . Here  R  is the mean distance between the centres of the Earth and the Sun, and  M S &nbs

Accepted Answer

T = 2 &#960; R 3 / 2 G M S For binary stars, T ' = 2 &#960; r 3 / 2 G M 1 + M 2 Put  M 1 = 3 M S , M 2 = 6 M S  and  r = 9 R T ' = 2 &#960; &#215; 9 3 / 2 &#215; R 3 / 2 G 3 M S + 6 M S = 2 &#960; &#215; 9 3 / 2 &#215; R 3 / 2 9 1 / 2 G M S = 9 2 &#960; R 3 / 2 G M S &#8756; &#160;&#160; T ' = 9 T Value of  n = 9 .

Gravitation — Previous Year Questions & Solutions