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.

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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)`

---

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)`

---

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`

---

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)`

---

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.

---

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

---

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.

---

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%`

---

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.

---

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

---

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))`

---

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.

---

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

---

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

---

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)`

---

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)`

---

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.

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