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Gravitation
Introduction
For making a body revolve, a centripetal force is required. In case of earth-sun system there exists a new
type of force called 'gravitational force' which is responsible for it. This force is always attractive in nature. In
this unit we will study in detail about this force and also calculate the velocities with which a body must be
projected from earth's surface. So that it never comes back to earth surface.
NEWTONS LAW OF GRAVITATION:
It states that every particle in the universe attracts every other particle with a force directly proportional to the
product of their masses and inversely proportional to the square of the distance between them.
Force is direct along the line joining the particles and towards other particle.
whre G => Universal gravitational constant.
=> Unit vector joining two particles.
° The gravitational force between two particles form an action reaction pair.
° If we consider a system which has both boies as its constituents then the net force becomes zero
.
° The magnitude of the forces acting is same on both bodies but direction is opposite.
Dumb Question: What if more than two particles are present.
Ans: In that case the principle of super position can be applied for summing the forces.
Illustration: Find the gravitational force of attraction on the point mass 'm' placed at O by a thin rod of mass
M and length L as shown in figure.
Solution: First we need to find the force due to an element of length dx. The mass of the element is dm =
dx.
So, dF =
The net gravitational force is
F=
Notice that when d >> L, we find F =
, the result for two point masses.
GRAVIATTIONAL FIELD:
Gravitation field is due to mass. It is defined as the region of space in which it interacts with other masses.
Another relevent quantity is gravitational field intensity.
° GRAVITATIONAL FIELD INTENSITY due to a mass 'm' at a distance 'r' is defined as the force acting on
unit mass kept at a distance r. It is a vector quantity and its direction is the direction along which te unit mass
has tendency to move.
Few gravitational fields :
° Due to a point mass
° Due to ring at an axial distance
° Field due to disk
inside = 0
° Due to thin spherical shell
outside =
inside =
° Field due to solid sphere
outside =
Why ??
Intensity due to a uniform circular ring at a point on its axis:
Figure shows a ring of mass M and radius. Let P is the point at a distance r from the center that is along
.
From the figure, AP =
Now field at P due to dm along
dE
has magnitude = dE =
Net Gravitational field E =
dry
Dumb Question: As the P is moved for away, how will the intensity change ?
Ans: for r >> R,
r2 + R2 r2
field varies inversely to the square of distance
however if r << R,
r2 + R2 R2
or E
r,
E=for very close distances, field varies directly with r.
• Intensity due to a uniform disc at a point on its axis:
Let mass of disc be M and radius be R and P be pt. m its axis where gravitational field is to be calculated.
Taking a small ring of thickness dx and radius x in the disc.
Man of the ring dm =
field at P due to ring is
in term of
E
• Field due to a solid sphere:
Let the mass of sphere is M and its radius is R,
Case I: External point
Choosing a thin shell of mass dm
then dE =
[for a point outside hollow shell, it acts as a point mass ... at centre of shell]
Case II: Field at internal point:
P is inside the sphere. Taking a small thin spherical shell of radius x and thickness dx[x < d]
Field due to all such shell sun
All shells having radius greater than P do not contribute to the field
Graphical representation of fields:
(i) Variation of gravitational field due to aring as a faver of its axial distance.
(ii) Gravitational field due to a uniform sphere
• As can be seen that the Gravitational field faver is continuous every where but is not differentiable at R.
° Field due to uniform thin spherical shell
Case I: When point lies inside the spherical shell
menclosed = 0, [
it is a shell i.e. hallow sphere]
Case II: When point P lies outside the spherical shell
Thus for outide point, whole shell acts a point object located at the centre of sphere.
Graphical representation of field.
Dumb Question: How can the result I be justified physically ?
Ans:
We consider under whole shell to be made up of minute particles each of mass 'dm'. Then for any point P
inside, the force diagram will be as shown in adjucant figure. Now since section
is smaller to P, hence
force due to than will be stronger (as Force follows inverse square relation with distance] hence lesser
particles with stronger force compensates more particles with weater force an left side.
Illustration: Two concentric shells of masses M1 and M2 are situated as shown in figure. Find the force on a
particle of mass 'm' when the particle is located at (a) r = a (b) r = b (c) r = c. The distance r is measured
from the centre of the shell.
Solution: We know that attraction at an external point due to spherical shell of mass M is
internal point is zero.
So,
(a) For r = a, the point is external for both the shell; so
EA =
while at an
FA = mEA =
(b) For r = b, the point is external to the shell of mass M2 and internal to the shell of mass M1;
So,
EB =
+0
FB = mEB =
(c) For r = c, the point is internal to both the shells, so,
EC = 0 + 0 = 0
FC = mEC = 0
GRAVITATIONAL POTENTIAL:
Gravitational potential is defined for a point gravitational field. It is the work done by the external agent
against the gravitational field in bringing unit mass from infinity to taht point without acceleration.
If work done is W then V =
By the definition of potential energy u = W
or U = mV
In differential term
V=-
where
or E = Dumb Question: While defining potential, why the constraints of 'zero acceleration' been placed ?
Ans: Potential being defined for conservative forces, is actually the anount of work that has been done
against and conservative force must be perfectly balanced and hence comes in the constraint of 'zero
acceleration'.
GRAVITATIONAL POTENTIAL FOR FEW CASES:
• Gravitational potential at a point due to a point mass.
V=-
Why ??
Gravitational field due to a point mass is given by E = -
For all potential calculations, untill unless mentioned take zero of potential at infinity.
• Potential at a point due to a ring
at point P, it is given as
V=-
, M = man of ring
• Due to spherical shell
Case I: Point P lies outside the shell V = -
Case II: Point P lies inside the shell V = Graphical result can be shown as
• Potential due to homogenous solid sphere
Case I: when the point P lies outside the sphre
V=Case II: When the point P lies inside the sphere
V=Graphical representation
Why ??
• Field due to a ring.
Let M be the mass and R be radius of a thin ring taking a small element of the eing and heating it as a point
mass, the potential due to it is
GRAVITATIONAL POTENTIAL ENERGY:
Gravitational force is conservative and have potential energy is defined for it.
in previous section it was seen
V
Potential at a point where 'm' is kept, due to all masses present.
or mV = U=-
= Vm
For two particles potential energy is given by Vm 2 = Dumb Question: We know that potential energy is a relative term, however the formula above gives different
inter pretation.
Solution: The above formula is for the case where the objects potential energy is calculated relative to zero
point of potential energy.
The General formula should be
Uf - Ui = where Uf and Ui represents final and initial potential energy (w.r.t. zero point) and r2 and r1 represents the
repective location of final and initial point.
Dumb Question: Is the force used in above formula external or gravitational ?
Solution: The force used in above formula is 'FORCE OF GRAVITATION'. If external force is given, then it
can be used by just dropping the -ve sign of the formula and using rest of the things as it is, but the motion
of the body must follow the constraight of 'ZERO ACCELERATION' if external force is being used using
'Fgravity' formula holds, irrecpective of acceleration.
ASSEMBLING A SYSTEM (POTENTIAL ENERGY OF A SYSTEM:
• If there is a system consisting of few objects, then the potential energy of complete system is same as the
work done in assembling the system (with initially nothing present).
Illustration: Three particles each of mass m are placed at the corner of an euilateral triangle of side d.
Calculate
(a) Work done in assembling the configuration.
(b) Work done on the system if the side of the triangle is changed from 'd' to '2d'.
Solution:
(a) Potential Energy of Complete System = U12 + U23 + U31
The work done in assembling the whole system will be same as this potenial energy.
(b) New potential energy = Uf = -
Work done in formation of new style = +
Work done by external force =
(a) Potential Energy of Complete System = U12 + U23 + U31
VARIATION IN ACCELERATION DUE TO GRAVITY:
(a) With Altitude, g = g0(1 -
(b) With depth, g = g0(1 -
)
)
Why ??
With Altitude:
At the surface of the earth
g=
for height 'h' above the surface of the earth,
g' =
for h << R. Using Binomial approxiation.
or g' = g
[note: formula valid only for h << R]
With depth:
Suppose body is at depth d assuing density of eargth to remain same throughout.
From Shell's theorem, the body at this depth will only experience the force because of sphere, it is enclosing
man of that sphere =
force on unit mass due to it = g
Illustration: Calculate the acceleration due to gravity at the surface of Mars, if its man is 1/10 th that of the
earth and diameter 1/1.88 that of erth, (gE = 9.8 m/s2).
Solution: We know g =
gM = 0.35344 gE
= 0.35 x 9.8
= 3.48 m/s2
ESCAPE SPEED:
It is the minimum speed with which a body must be projected from the surface of a planet. So that it escapes
from the gravitational field of the planet (i.e. it comes to rest at infinity)
Ve =
Ve = 11.2 km/s
Why ??
Potential energy of the body at surface of planet = -
If it is launched of planet =
mVe2
Total mechanical energy of system =
mVe2 At infinity potential energy = 0 (as
is assumed ....)
Kinetic energy = 0 (as body comes to rest hypothetically at infinity
Total mechanical energy now = 0
From conservation of mechanical energy
mVe2 = 0 or Ve =
It is important to not that escape speed is independent of the mass and direction of projection of the body.
Illustration: The masses and radii of the earth and the moon are M1, R1 and M2, R2 respectively. Their
centres are at a distance d apart. What is the minimum speed with which a particle of man m should be
projected from a point midway between the two centres so as to escape to infinity ?
Solution: Potential energy of m which is midway between M1 and M2 is
U = m(v1 + v2) = m
Let v be the requiyed speed, then
Total mechanical energy
+
mv2
Total mechanical energy of mechanical energy = 0
From conservation of mechanical energy
+
mv2 = 0
or v =
MOTION OF SATELLITE:
• ORBITAL VELOCITY:
The velocity of a satellite in its orbit is called orbiatry velocit. If V 0 is orbittal velocity then
(for centripetal force balance)
=> v0 =
where h => height above ground
Dumb Question: What is the direction of this velocity ?
Ans: Orbitary velocity the velocity tangential to the path of motion of satellite. It is also independent of the
mass of satellite.
• ENERGY OF SATELLITE
P.E. of a satellite = -
K.E. of a satellite =
mV02 =
m
=
Total mechanical energy of satellite = P.E. + K.E.
=-
+
=Dumb Question: What is the significance of -ve total mechanical energy ?
Ans: Negative energy means that the satellite is bound to the central body by an attractive force and energy
must be supplied to remove it from the orbit to infinity.
• GEO-STATIONARY SATELLITE:
If there is a satellite rotating from west to east and havig same angular velocity a earths rotation, then such a
satellite appers stationary from the orbit is called GEO-STATIONARY.
The orbit Geostationary satellite is called 'PRKING ORBIT'.
• TIME PERIOD OF SATELLITE:
T=
where r -> radius of orbit
M -> man of central planet
How ??
T=
or T2 =
KEPLER'S LAWS:
Kepler gave three empirical laws for the motions of planet around a central star(Sun)
(i) Each planet moves in an elliptical orbit, with at one focus of the ellipse.
(ii) The radius vector, drawn from the sum to a planet, sweeps out equal areas inequal time i.e. its a real
velocity ( or area swept out by it per unit time) is constant.
= constant,
where A -> Area swept by vector.
L -> Angular momentum of planet
m -> man of planet
=> L = constant
(iii) The square of the planet's time period is proportional to the cube of te semi-major axis of its orbit.
T2 a3
Precisely T =
where M5 = man of sun
• Important interpretation of IInd law
Let the area swept be shown in the adjucant figure.
dA =
(base)(height) =
=
(r)(v sin
(SP)(P'M)
)dt
or
Since
= constant
=>
=> L = constant. Now
=0
Now external torque acts on the system.
= constant
Dumb Question: What path will a particle follow. If give velocity v such that
(i) v < v0 where v0 = orbitary velocity.
(ii) v0 < v < ve ve = escape velocity
(iii) ve < v
Ans: (i) The path of particle is elliptical and it may or may not fall back on earth. If elliptical path crosses the
earth, then it will fall back else not.
(ii) For V0 < v < ve, the particle will again follow elliptical path, but this time the point of projection is the
nearest point from the focus, in the locus.
(iii) v > ve, The path followed is hyperbolic while for v = ve the path is parabolic.
Illustration: The mean distance of mass from the sun is 1.524 times that of the earth from the sun. Find the
number of years required for mars to make on revolution about the sun.
Solution: For planets revolving around the sun T2 r3
[T1 : time-period of mars, T2 : time-period of earth]
= (1)(1.524)3/2 = 1.88 years
PROBLEMS:
Easy:
1. Complete the mass and density of the moon if acceleration due to gravity on its surface is 1.62 m/s 2 and
radius is 1.74 x 106 (G = 6.67 x 10-11)
Solution: We know that
g=
or M =
= 7.35 x 1022Kg
Now P =
= 3.3 x 103 Kg/m3
2. What will be the acceleration due to gravity on the surface of the moon if its radius 1/4 th the radius of the
earth and mass 1/80th the mass of the earth. What will be the escape speed on the surface of the moon if it
is 11.2 km/s on the surface of the earth.
Solution: As on the surface of planet
g=
, we have
gM = gE/5 = 9.8/5 = 1.96 m/s
further as escape speed = ve =
vM =
, So
= 2.5 km/s
3. The orbiatry velocity of an artificial satellite in a circular orbit just above earth's surface is v 0. What is the
orbitary velocity for a satellite orbitary in a circular orbit at an attitude of half of earth's radius.
Solution: v0 =
if h = 0
if h = R/2
v0 =
v0 =
=
v0
4. Find the gravitational force of attraction between a uniform sphere of mass M and a uniform rod length 'l'
and mass 'm' oriented as shown in figure.
Solution: Since the sphere is uniform its entire mass may be considered to be concentrated at its centre.
The force on elementary mass dm is
dF =
But dm =
dx
(thickness neglected)
F=
=
5. An orbitary satellite is moving in a circular orbit around the earth, with a speed equal to half the magnitude
of escape velocity from earth.
Ditermine
(a) The height of satellite above earth's surface.
(b) If the satellite suddenly stops. Find the speed with it hit the earth's surface after falling down.
Solution: (a) ve =
and v0 =
Now v0 =
or 2(R + h) = 4R
=> h = R
(height of satellite above ground)
(b) If the satellite is stopped in orbit, the kinetic energy is zero and
Potetial Energy = -
Total Energy = let velocity on reaching earth is 'v'
Potential Energy on the earth's surface = From conservation of mechanical energy
-
or
=-
+
mv2
v=
6. Suppose the earth increases its speed of rotation. At what new time period will the weight of a body in the
equator becomes zero ?
(Take g = 10 m/s2 and RE = 6400 km)
Solution: The weight will become zero when g' = 0
also g' = g - Rw2
=> g - Rw2 = 0
=> w =
T = 1.4 hr
7. A particle of mass 1 Kg is kept on the surface of a uniform sphere of man m = 20 Kg and radius r = 1.0 m.
Find the work to be done against the gravitational force between then to take the particle away from the
sphere.
Solution: Potential at the surface of sphere,
v== - 1.334 x 10-9 J/Kg
i.e. 1.334 x
J work is obtained to bring a man of 1 Kg from
to the surface of sphere. Hence, the same
amount of work will have to be done to take the particle away from the surface of the sphere.
Thus w = 1.334 x 10-9 J
10-9
8. A planet of mass m revolves in elliptical orbit around the sun. So that its maximum and minimum distance
from the sun are equal to ra and rp respectively. Find the angular momentum of this planet relative to the
sun.
Solution: Using conservation of angular momentum
mvprp = mvara
[As velocities or
.... to the radius vector at apogee and perigee
=> vprp = vara
Using conservation of energy
By solving the above equations,
vp =
L = mvprp = m
MEDIUM:
1. Two bodies of masses M1 and M2 are placed at a distance d a point. What is the potential at the position
where the gravittional field due to them is zero ?
Solution: Let the field be zero at a point distance x from M1
Potential at this point due to both the masses will be
2. A rocket starts vertically with speed v0. Show that its speed v at a height is given by v02 - v2 =
,
where R is the radius of earth. Hence deduce the maximum height reached by the rocket fired with a speed
of 900 of escape velocity.
Solution: Kinetic Energy on surface of earth =
Potential Energy on surface of earth = -
mv02 -
Total energy =
Kinetic Energy at a height h =
mv2
Potential Energy at this height = -
Total energy =
mv2 By the principle of conservation of energy
mv2 -
=
mv02 -
mv02
(v02 - v2) =
But GM = gR2
(v02 - v2) =
v02 - v2 =
For maximum height v = 0
v0 = 90% of escape speed = 0.9
0.81R + 0.81hmax = hmax
0.19hmax = 0.81R
hmax =
= 4.26R
3. Two satellites A and B of equal masses, move in the equitorial plane of earth, close to the earth's surface.
Satellite A moves in the same direction as that of the rotation of earth while satellite B moves in the opposite
direction. Determine the ratio of the kinetic energy of B to that of A in the reference frame fixed to earth (g =
9.8 m/s2)
Solution:
Velocity of A with respect to earth =
- wR
Velocity of B with respect to earth =
+ wR
= 1.265
where T = 24 x 60 x 60 s
R = 6400 x 103 m
HARD:
1. An artificial satellite of mass m of a planet of mass M, resolves in a circular orbit whose radius is n times
the radius R of the planet. In the process of motion, the satellite experiences a slight resistance due to
cosmic dust. Assuming resistance force on satellite depends on velocity as F = av 2 where a is constant.
Calculate how long the satellite will stay in orbit before it falls onto the planet's surface.
Solution: Total energy in circular orbit of radius r,
E=-
Rate of change of energy =
= Fresistiv. v = av2.v = av3
also gravity provides the centripetal force
Now
2. Find the maximum and minimum distances of the planet A from the sun S if at a certain moment of time it
was a distance r0 and travelling with the velocity vector being equal to
Solution:
.
At minimum and maximum distance velocity of satellite makes on angle of 90 0 with radius vector.
Applying conservation of angular momentum
mv0r0 sin
= mvr
By energy conservation
Solving equation (i) and (ii) we get two values of r, one is maximum distance another is minimum distance.
rmax =
rmin =
where
Key words:
• Gravitational Force.
• Gravitationl Field.
• Gravitational Potential.
• Escape Speed.
• Orbittal Speed.
• Areal Velocity.
• Parking Orbit.