The following question will be simple if you have a clear idea of the potential energy in a gravitational field:
Two bodies of masses m1 and m2 are initially at rest and and at infinite distance apart. When a very feeble momentary push is given to one of them, they move towards each other because of the gravitational force between them. When the separation between them is ‘r’, their relative velocity of approach is (if G is the gravitational constant)
(a) [2Gr(m1+m2)]½ (b) [2Gr(m1–m2)]½ (c) [1/2Gr(m1+m2)]½
Two bodies of masses m1 and m2 are initially at rest and and at infinite distance apart. When a very feeble momentary push is given to one of them, they move towards each other because of the gravitational force between them. When the separation between them is ‘r’, their relative velocity of approach is (if G is the gravitational constant)
(a) [2Gr(m1+m2)]½ (b) [2Gr(m1–m2)]½ (c) [1/2Gr(m1+m2)]½
(d) [2G (m1+m2)/r]½ (e) [2G(m1 – m2)/r]½
Imagine that initially the masses are at a distance ‘r’ apart. The gravitational potential energy of the system then is –Gm1m2/r. If the separation between the masses is to be increased to infinity, kinetic energy equal to Gm1m2/r is to be supplied to the system. (At infinite separation, kinetic energy as well as potential energy is zero). When the masses are allowed to approach, they lose potential energy (which becomes negative) and gain kinetic energy and at a separation ‘r’, the kinetic energy is +Gm1m2/r.
If v1 and v2 are the velocities of the masses, we have
½ m1v1 2 + ½ m2v2 2 = Gm1m2/r.
Since the momenta are equal (in magnitude), m1v1 = m2v2 so that v2 = m1v1/m2.
Substituting this in the energy equation above, v1 = [2G/(m1 + m2)r]½ ×m2.
Similarly, v2 = [2G/(m1 + m2)r] ½ ×m1.
Since the masses are moving in opposite directions, their relative velocity is
v1 + v2 =[2G/(m1 + m2)r]½ × (m2 + m1) = [2G(m1+m2)/r]½
[The solution to this problem has been edited. Thank you Dr. Ravikrisnan, for pointing out the mistake in the solution which I had posted yesterday].
Imagine that initially the masses are at a distance ‘r’ apart. The gravitational potential energy of the system then is –Gm1m2/r. If the separation between the masses is to be increased to infinity, kinetic energy equal to Gm1m2/r is to be supplied to the system. (At infinite separation, kinetic energy as well as potential energy is zero). When the masses are allowed to approach, they lose potential energy (which becomes negative) and gain kinetic energy and at a separation ‘r’, the kinetic energy is +Gm1m2/r.
If v1 and v2 are the velocities of the masses, we have
½ m1v1 2 + ½ m2v2 2 = Gm1m2/r.
Since the momenta are equal (in magnitude), m1v1 = m2v2 so that v2 = m1v1/m2.
Substituting this in the energy equation above, v1 = [2G/(m1 + m2)r]½ ×m2.
Similarly, v2 = [2G/(m1 + m2)r] ½ ×m1.
Since the masses are moving in opposite directions, their relative velocity is
v1 + v2 =[2G/(m1 + m2)r]½ × (m2 + m1) = [2G(m1+m2)/r]½
[The solution to this problem has been edited. Thank you Dr. Ravikrisnan, for pointing out the mistake in the solution which I had posted yesterday].
Now, consider the following question, which will be simple if you have a clear idea of the velocity of escape in a gravitational field:
The surface value of acceleration due to gravity on a planet of radius ‘R’ is ‘g’. If a body of mass ‘m’ at infinite distance from the planet is given a feeble momentary push so that it moves towards the planet, what will be its kinetic energy when it strikes the surface of the planet? (Assume that the gravitational field of the planet only is significant and the effect of atmosphere on the motion of the body is negligible).
(a) Infinite (b) mgR2 (c) ) mgR3 (d) 2mgR (e) mgR
The body will strike the surface of the planet with a velocity equal to the escape velocity which is √(2gR). The kinetic energy with which it will strike the surface is therefore ½ m[√(2gR)]2 = mgR.
The surface value of acceleration due to gravity on a planet of radius ‘R’ is ‘g’. If a body of mass ‘m’ at infinite distance from the planet is given a feeble momentary push so that it moves towards the planet, what will be its kinetic energy when it strikes the surface of the planet? (Assume that the gravitational field of the planet only is significant and the effect of atmosphere on the motion of the body is negligible).
(a) Infinite (b) mgR2 (c) ) mgR3 (d) 2mgR (e) mgR
The body will strike the surface of the planet with a velocity equal to the escape velocity which is √(2gR). The kinetic energy with which it will strike the surface is therefore ½ m[√(2gR)]2 = mgR.
No comments:
Post a Comment