The minimum work done to carry an object in a gravitational field is the work to be applied in order for the difference of kinetic energy be zero.
Transcript
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Gravitation
Gravitation- By Aditya Abeysinghe
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Newton’s law of universal gravitation
M
F
F
d
Newton expressed that the force of attraction between any two objects is
directly proportional to their product of masses (mM) and inversely proportional
to the square of the distance between the two objects
F = GMm / d2 G = 6.67 × 10-11 Nm2 kg-2
Gravitation- By Aditya Abeysinghe
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Gravitational field strength (E)
Gravitational
field strength is
defined as the
force per unit
mass
F = mg
r
F = GMm/ r2
E = F/m = GM/ r2
E = GM/ r2
F = GMm/ r2
Therefore,
E = g = GM/ r2
mg = GMm/ r2
Therefore, g
= GM/ r2
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Expressing g using the density of a planet
V= (4/3)πr3 ρ= M/V
M= (4/3)πr3ρ
g= GM/r2 = G × (4/3)πR3ρ
R
r2
When, r=R
That is,
r
g= (4/3)πRρ
From the above derivation,
gαρ
g α R and g α 1/r2
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Although we assume that the Earth is a uniform planet, its
density varies from about 2800 kgm-3 in the crust to about
9700 kgm-3 at the core.
Due to this, the value of g increases with depth for some
distance below the surface.
Variation of acceleration of free-fall or field strength with
distanceg’
Assuming uniform density inside the Earth
g
At the surface of the Earth
Inside
Outside
Inverse square law ( g α 1/r2 )
0
rE
Gravitation- By Aditya Abeysinghe
r
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Variation of acceleration due to gravity with
height
At a height h above the Earth’s surface,
gh = GM/ (R + h)2 ; R- Earth’s radius.
h
R+h
R
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Variation of acceleration of gravity with
depth
Consider a body at a depth d below the surface of the
earth. Thus, r= R- d .
R
d
r=R-d
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It can be shown that the net attraction on the body is due to
the mass of the earth contained in the sphere with radius r. If
M’ be the mass of this part, then the acceleration due to gravity
is
gd = GM’/ r2 . If we assume that the earth has uniform density
ρ, then
M’ = (4/3) πr3 ρ
Therefore, gd = G ((4/3) πr3 ρ)/ r2 = (4/3)Gπrρ
Therefore, gd = (4/3)Gπρ(R-d)
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Variation of g on the surface of the earth
Although we assume g on earth is a constant value at any place, practically the value
of g varies at various points on earth. The value of g is maximum at the poles and
then decreases towards the equator and is thus minimum at the equator. Two
factors contribute towards this variation:
1. Shape of the earth- Earth has a ellipsoidal shape which bulges at the poles and
flattens at the poles. Therefore, the polar radius is less than the equatorial
radius. Therefore from the formula of g
(g= GM/r2 ), it should be evident
that g α 1/r2 and thus g should be greater at the poles
2. Rotation of the earth- Due to earth’s axis rotation bodies on earth experience
centripetal acceleration. Thus the attraction force of an object on earth is
actually the sum of the weight of the body and the force exerted by the
centripetal force.
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Thus, total force F,
F= Weight of the object + Centripetal force
F= W + mRω2 (Where ω is the angular velocity of the earth)
W= F - mRω2
mg= GMm/R2 – mRω2
g = g0 – Rω2
At an inclination of λ from the equator (latitude λ) , it can be
Axis of rotationshown that,
g = g0 – Rω2Cos2λ
λ
• At poles, λ= 90° , Thus, g= g0
• At the equator, λ=0, thus, g=g0 – Rω2
Thus, it should be clear that the value of g increases as a body gradually
moves from the equator to the poles with the variation of λ
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Acceleration due to gravity of different
planets
Since different planets have different masses and radii, the
values of g due to their gravitational fields are different.
R1
R2
M1
g1 = GM1 / R12
M2
g2 = GM2 / R22
g1 / g2 = (M1 / M2) (R2 / R1 )2
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Gravitational Potential Energy
DefinitionGravitational Potential Energy of a body at a point in a
gravitational field is the work done by an external agent in
moving the body from infinity to that point.
Thus, potential energy(U) at any point is defined as,
U = -GMm/ r
Actually this is the potential energy of the Earth-Mass system.
Gravitational potential energy is always negative because the
work is always done by the gravitational field(not by the body in
the field)
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Thus the gravitational potential energy of a body at the earth’s surface is, U1 = - GMm/ R .
The gravitational potential energy of a body raised to height of h is, U2 = -GMm/ (R + h)
Thus the work done by the earth’s gravitational field to carry this object to a height of h is,
W= U2 - U1
W= [-GMm/ (R + h)] – [- GMm/ R ]
W= -GMm { 1/ (R + h) – 1/R }
W= GMmh/R(R + h). Since h<<R, it can be assumed that R + h ≈ R. Thus, R(R +h) can be written
as R2 .
Therefore, W= GMmh/ R2 OR W=(GM/R2 ) mh.
However, we know that g= GM/ R2 . Therefore,
W=mgh, which is the potential energy of an object raised to a height of
h
as described in
basic mechanics.
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Note:
Though the theoretical zero of potential
energy is infinity, in most problems, for our
convenience, we take the zero point of
potential energy as the earth’s surface.
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Gravitational potential at a point
It is the work done in moving a unit mass from infinity to that
point.
Thus, the gravitational potential at a point can be described
as
V = -GM/r , as the value of m in the gravitational potential
energy expression is equal to 1(unit mass).
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Escape Velocity
It is the minimum velocity with which an object should be
ejected from the planet so that it does not return back.
That is it goes to infinity with zero velocity.
The energy needed for such a scenario can be found by
the energy conservation principle.
(K.E. + P.E. )r=R = (K.E. + P.E.)r=infinity
½ mve2 + (-GMm/R) = 0 + 0
Ve – Escape
Therefore, ve = √(2gR)
Velocity
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Motion of planets and satellites
Consider a satellite of mass m moving around a planet of
mass M in a circular orbit radius r. If
V0 is the orbital velocity, then
mV02 / r = GMm/r2 , as the force of attraction is the
centripetal force.
Therefore V0 = (GM/r)½ . Thus, V0
= √gR
For earth, V0 ≈ 8km/s.
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Time period of revolution
T = 2π/ω = 2πr/ V0
T = 2πr (r/(GM))½
Therefore, T2 = 4π2r2 (r/(GM))
Therefore, T2 = 4 π2r3 /GM
Thus, T2 α r3
Therefore, T12 = r13
T22 r23
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Energy of a satellite
The total energy of an orbiting satellite can be written as the
sum of kinetic energy and the gravitational potential energy.
Therefore, E= EK + EP
E= ½mv02 + (-GMm/r)
= GMm/2r - GMm/r , as V02 = GM/r
= -GMm/r {1-½ }
Thus,
E = -GMm/2r
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Geostationary Satellite
(Parking Orbits)
An earth satellite is so positioned that it appears stationary
to an observer on earth. This is because of its 24h time
period of revolution as the earth. The satellite rotates at the
same velocity the earth rotates about its axis. Thus the
observer on earth neither feels the rotation of the earth nor
the rotation of the satellite.
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Relation between field strength and
potential gradient
Suppose a mass in a gravitational field is taken from point A to
point B through a small distance Δr, then,
Work done per kg= Force per kg × Δr
Force per kg= Average field strength in the A-B
region.
Therefore, work done= EG × Δr = ΔV (ΔV- potential energy change
from A to B)
EG = ΔV / Δr .
Therefore, Field Strength = Potential Gradient
But, potential decreases when r increases.
So, E = - dV/dr
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Equipotential Surfaces
The gravitational potential outside a spherical mass M at a point distance r from its
center is given by V= -GM/r
So, all points at a distance r have the same potential.
These points lie on a sphere of radius r. So, we call this sphere an equipotential
surface.
No work should be done to take
masses within the same surface.
But work should be done to take
masses from one surface to
another
Equipotential
surfaces
Gravitation- By Aditya Abeysinghe
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