Physics Definition And Proof Based Problems

### Definition And Proof Based Problems

Definition And Proof Based Problems
Q 3174078856

Derive a relation for work done in a gravitational field. Using it, (i) find potential difference between a pair of points. (ii) express whether gravitational force is conservative or non-conservative.
1976
Solution:

The gravitational force of attraction between M and m when x is the distance between their centres is given by

F = (GMm)/(x^2)

Suppose the body is moved through a distance
dx, therefore, work done is given by,

dW= Fdx= (GMm)/(x^2) dx

When the body is brought from infinity to
some distance r,

we write int dw = int_(x= oo)^(x = r) (GMm)/(x^2) dx

or W = GNM [(-1)/(x)]_(oo)^(r)

= - GMm[1/r- 1/oo ] =(-GMm)/(r)
This amount of work done is the change in
the potential energy of the body.

P.E U = - (GMM)/(r)

Gravitational potential

V = U/m = (-GM)/(r)
The general expression for gravitational
potential due to the earth (mass M) at

(i) distance r is, v = (-GM)/(r)

Potential at a point A (r_a) = - (GM)/(r_a)

Potential at a point B (r_b) = - (GM)/(r_b)

:. Difference in potential between the points

= - GM(1/r_a-1/r_b)

(ii) Since work done against gravitational force is (a) independent of path and dependent only on the initial and final points and (b) the work done in a closed path is zero, it is a conservative force.
Q 3124378251

Define gravitational potential energy of a body. Derive an expression for the gravitational potential energy of a body of mass 'm' located at a distance 'r' from the centre of the earth.

Solution:

Gravitational potential energy. The work done in carrying a mass 'm' from infinity to a point at distance r is called gravitational potential energy.

G.P.E = - (GMm)/(r)

i.e.,G.P.E = Mass x Gravitational potential It is a scalar quantity measured in joule. Negative sign means that the mass is bound to M.
The gravitational force of attraction between M and m when x is the distance between their centres is given by

F = (GMm)/(x^2)

Suppose the body is moved through a distance
dx, therefore, work done is given by,

dW= Fdx= (GMm)/(x^2) dx

When the body is brought from infinity to some distance r,

we write int dw = int_(x= oo)^(x = r) (GMm)/(x^2) dx

or W = GNM [(-1)/(x)]_(oo)^(r)

= - GMm[1/r- 1/oo ] =(-GMm)/(r)

This amount of work done is the change in the potential energy of the body.

P.E U = - (GMM)/(r)
Q 3154178954

What is the difference between gravitational potential and gravitation potential energy. Derive an expression for gravitational potential energy of a body.

Solution:

The gravitational force of attraction between M and m when x is the distance between their centres is given by

F = (GMm)/(x^2)

Suppose the body is moved through a distance
dx, therefore, work done is given by,

dW= Fdx= (GMm)/(x^2) dx

When the body is brought from infinity to
some distance r,

we write int dw = int_(x= prop)^(x = r) (GMm)/(x^2) dx

or W = GNM [(-1)/(x)]_(prop)^(r)

= - GMm[1/r- 1prop ] =(-GMm)/(r)
This amount of work done is the change in
the potential energy of the body.

P.E U = - (GMM)/(r)

Gravitational potential

V = U/m = (-GM)/(r)
The general expression for gravitational
potential due to the earth (mass M) at

v = (-GM)/(r)

Work done to carry any mass m is stored
as P.E. in it at that position and is

P.E = - (GMm)/(r)
Q 3164478355

Derive an expression for the orbital velocity of a satellite in the orbit. Reduce it to an orbit close to the surface of earth. How is it related to escape velocity ?

Solution:

In an orbit of radius r, a satellite of mass m
moves round a planet of mass M. Then,

(mv_(0)^(2)/(r)) = (GMm)/(r^2)

v_0 = sqrt(GM)/(r) = sqrt(GM)/(R+h)

where h is the height at which the satellite is from the surface.

For close to earth orbits, h = 0

v_0 = sqrt(GM)/(R) = sqrt(gR)

Since v_e = sqrt(2gR.,) v_0 = (v_e)/sqrt2
Q 3114456359

What is escape velocity. Derive an expression for the same.

Solution:

The minimum velocity required to escape from the gravitational force of earth is called escape velocity. Total energy is the sum of P.E. and K.E.

T.E = - (GMm)/(R) +1/2mv^2

To escape K.E. should be greater than P.E., i.e.,

1/2 mv^2>= - (GMM)/(R)

v_e = sqrt(2Gm)/(R)= (sqrt2gR)
Q 3154578454

(i) Define escape velocity.
(ii) Derive expression for the escape velocity of an object from the surface of a planet.
(iii) Does it depend on location from where it is projected ?[

Solution:

(i) The minimum speed required for an object to reach infinity (i.e., to get escape from earth) is called escape velocity.

(ii) 1/2m(v_(1)^(2))_e = (GmM_P)/(h+R_P)

[where M_P = Mass of planet R_P = Radius of planet]

If the object is thrown from surface of
a planet h = 0, we get

(v_i)_e = sqer(2GM_P)/(R_P)

but g = (GM_P)/(R_(P)^(2) we get

[ (v_i)_e = sqrt(2gR_P)

(iii) Depends on location as 'g' varies with
location, as most of the celestial bodies
are not perfectly spherical.
Q 3104078858

State Kepler's laws on planetary motion. Explain the way the three laws can be proved.

Solution:

Kepler's Laws of planetary motion : l.All the planets move in elliptical path with sun at their foci. 2.The line joining the sun and the planet sweeps out equal areas in equal intervals of time. 3.The square of the time period of revolution is proportional to the cube of the semi-major axis of the elliptical orbit. T^2 propa a^3. If r_1 and r_2 are the shortest and the longest

distances of the planet from the sun the

semi - major axis is given by (r_1+r_2)/(2)
1st law is proved from angular momentum
conservation in a circular path.
Since L = 2m xx aerial velocity, for equal
time, area swept will be same. So 2nd law
is proved.

From F prop 1/r^2 (mv^2)/(r) = F and T = (2pir)/(v) one can
prove T^2 prop r^3 thereby proving 3rd law
Q 3114578450

(a) According to Kepler's second law, the radius vector to a planet from the sun sweeps out equal areas in equal interval of time. The law is consequence of which conservation law?
(b) State Kepler's third law.

Solution:

(a) Law of conservation of angular momentum.
(b) Kepler's third law is also known as law of periods, the square of the period of revolution of a planet around the Sun is proportional to the cube of semi-major axis of elliptical orbit.
Q 3174878756

State Newton's law of Gravitation. Find the percentage decrease in the weight of the body when taken to a height of 16 km. above the surface of the earth. Radius of the earth is 6400 km.

Solution:

It states that every body in the universe attracts every other body with a force which is directly propotional to the product of their masses and inversely propotional to square of distance between them.

F = (Gm_1m_2)/(r^2)

m_1 and m_2 ->t mass of two bodies r-> distance between two bodies. The acceleration due to gravity at a height 'h' above the surface of the earth is

g = g (1 - (2h)/R)

g - g' = (2hg)/(R)

(mg-mg)/(mg) xx 100 = (g-g)/(g) xx100

= (2h)/(R) xx 100

= (2xx16)/(6400) xx 100 = 0.5%
Q 3144056853

State Kepler's laws of planetary motion and deduce Newton's Law of gravitation from them.

Solution:

(i) The planets including earth, go around the sun in elliptical m:bits.
(ii) The line joining the Sun and the planet sweeps equal areas in equal intervals of time.
(iii) The square of the time period of revolution is directly proportional to the cube of the semi-major axis of the elliptical orbit.

Since T^2 prop r^3 , we have,
(2pir)/(v) prop r^3

v^2 = 4pir^2 (r^2/r^3)= (4pi^2)/(r)

(mv^2)/(r) = (4mpi^2)/(r^2)

The centripetal force (mv^2)/(r) is caused by
M - earth on the planet of mass m.
Thus, Falpha (Mm)/(r^2)
It is the Newton's Universal Law of Gravitation.
Q 3104078858

State Kepler's laws on planetary motion. Explain the way the three laws can be proved.

Solution:

Kepler's Laws of planetary motion : l.All the planets move in elliptical path with sun at their foci. 2.The line joining the sun and the planet sweeps out equal areas in equal intervals of time. 3.The square of the time period of revolution is proportional to the cube of the semi-major axis of the elliptical orbit. T^2 propa a^3. If r_1 and r_2 are the shortest and the longest

distances of the planet from the sun the

semi - major axis is given by (r_1+r_2)/(2)
1st law is proved from angular momentum
conservation in a circular path.
Since L = 2m xx aerial velocity, for equal
time, area swept will be same. So 2nd law
is proved.

From F prop 1/r^2 (mv^2)/(r) = F and T = (2pir)/(v) one can
prove T^2 prop r^3 thereby proving 3rd law
Q 3164378255

Prove that gravitational potential difference is the work done in carrying a unit mass from one point to another.

Solution:

Gravitational potential at A = -(GM)/(r_a)

Gravitational potential at B = -(GM)/(r_b)

Difference in potential = -GM (1/(r_b) - 1/(r_a))

Work done in carrying a unit mass from A to B is,

W = int_(r_a)^(r_b) (GM)/(x^2)

W= - GM |1/x|_(r_a)^(r_b)

= -GM (1/(r_b) - 1/(r_a))
Q 3174078856

Derive a relation for work done in a gravitational field. Using it, (i) find potential difference between a pair of points. (ii) express whether gravitational force is conservative or non-conservative.
1976
Solution:

The gravitational force of attraction between M and m when x is the distance between their centres is given by

F = (GMm)/(x^2)

Suppose the body is moved through a distance
dx, therefore, work done is given by,

dW= Fdx= (GMm)/(x^2) dx

When the body is brought from infinity to
some distance r,

we write int dw = int_(x= oo)^(x = r) (GMm)/(x^2) dx

or W = GNM [(-1)/(x)]_(oo)^(r)

= - GMm[1/r- 1/oo ] =(-GMm)/(r)
This amount of work done is the change in
the potential energy of the body.

P.E U = - (GMM)/(r)

Gravitational potential

V = U/m = (-GM)/(r)
The general expression for gravitational
potential due to the earth (mass M) at

(i) distance r is, v = (-GM)/(r)

Potential at a point A (r_a) = - (GM)/(r_a)

Potential at a point B (r_b) = - (GM)/(r_b)

:. Difference in potential between the points

= - GM(1/r_a-1/r_b)

(ii) Since work done against gravitational force is (a) independent of path and dependent only on the initial and final points and (b) the work done in a closed path is zero, it is a conservative force.
Q 3104256158

State Newton's law of gravitation in vector form.

Solution:

The force of attraction between a pair of masses m_1 and m_2 separated by a length 'r' is given by.

vecF_12 = - (Gm_1m_2)/(r^2) hatr_(21)
[F_12 -> force on 1 due to 2]

hatr_(21)-> Points from 2 to 1
-ve sing shows that the force is attractive
Q 3104078858

State Kepler's laws on planetary motion. Explain the way the three laws can be proved.

Solution:

Kepler's Laws of planetary motion : l.All the planets move in elliptical path with sun at their foci. 2.The line joining the sun and the planet sweeps out equal areas in equal intervals of time. 3.The square of the time period of revolution is proportional to the cube of the semi-major axis of the elliptical orbit. T^2 propa a^3. If r_1 and r_2 are the shortest and the longest

distances of the planet from the sun the

semi - major axis is given by (r_1+r_2)/(2)
1st law is proved from angular momentum
conservation in a circular path.
Since L = 2m xx aerial velocity, for equal
time, area swept will be same. So 2nd law
is proved.

From F prop 1/r^2 (mv^2)/(r) = F and T = (2pir)/(v) one can
prove T^2 prop r^3 thereby proving 3rd law
Q 3124378251

Define gravitational potential energy of a body. Derive an expression for the gravitational potential energy of a body of mass 'm' located at a distance 'r' from the centre of the earth.

Solution:

Gravitational potential energy. The work done in carrying a mass 'm' from infinity to a point at distance r is called gravitational potential energy.

G.P.E = - (GMm)/(r)

i.e.,G.P.E = Mass x Gravitational potential It is a scalar quantity measured in joule. Negative sign means that the mass is bound to M.
The gravitational force of attraction between M and m when x is the distance between their centres is given by

F = (GMm)/(x^2)

Suppose the body is moved through a distance
dx, therefore, work done is given by,

dW= Fdx= (GMm)/(x^2) dx

When the body is brought from infinity to some distance r,

we write int dw = int_(x= oo)^(x = r) (GMm)/(x^2) dx

or W = GNM [(-1)/(x)]_(oo)^(r)

= - GMm[1/r- 1/oo ] =(-GMm)/(r)

This amount of work done is the change in the potential energy of the body.

P.E U = - (GMM)/(r)
Q 3154178954

What is the difference between gravitational potential and gravitation potential energy. Derive an expression for gravitational potential energy of a body.

Solution:

The gravitational force of attraction between M and m when x is the distance between their centres is given by

F = (GMm)/(x^2)

Suppose the body is moved through a distance
dx, therefore, work done is given by,

dW= Fdx= (GMm)/(x^2) dx

When the body is brought from infinity to
some distance r,

we write int dw = int_(x= prop)^(x = r) (GMm)/(x^2) dx

or W = GNM [(-1)/(x)]_(prop)^(r)

= - GMm[1/r- 1prop ] =(-GMm)/(r)
This amount of work done is the change in
the potential energy of the body.

P.E U = - (GMM)/(r)

Gravitational potential

V = U/m = (-GM)/(r)
The general expression for gravitational
potential due to the earth (mass M) at

v = (-GM)/(r)

Work done to carry any mass m is stored
as P.E. in it at that position and is

P.E = - (GMm)/(r)
Q 3174178956

What is escape velocity ? Obtain the expression for the escape velocity on earth. Why is it that there is no atmosphere on the moon ? Explain.

Solution:

The minimum velocity required to escape
from the gravitational force of earth is called

escape velocity. Total energy is the sum of P.E. and K.E.

T.E = (-GMm)/(R) + 1/2 mv^2

To escape, K.E. should be greater than P.E.,
i.e.,

1/2mv^2>= (GMm)/(R)

v_e = (sqrt2 (GM)/(R)) = sqrt(2gR)

The escape velocity from the moon's surface is about 2.38 km/see which is less than the r.m.s velocity of the air molecules. Thus all the molecules of gases have escaped from the surface of moon. So, there is no atmosphere.