Gravitation

Gravitation- By Aditya Abeysinghe 1

Gravitation


Newton’s law of universal gravitation

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

M F F

d


Gravitational field strength (E) F = GMm/ r2

E = F/m = GM/ r2

E = GM/ r2

F = GMm/ r2

Therefore, mg = GMm/ r2

E = g = GM/ r2 Therefore, g = GM/ r2

F = mg

r

Gravitational field strength is defined as the force per unit mass


Expressing g using the density of a planet

V= (4/3)πr3 ρ= M/VM= (4/3)πr3ρg= GM/r2 = G × (4/3)πR3ρ

When, r=R g= (4/3)πRρThat is, From the above derivation, g α ρ g α R and g α 1/r2

R

r2

r


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 distance-

Inside Outside

g’

g

0 rEr

Assuming uniform density inside the Earth

Inverse square law ( g α 1/r2 )

At the surface of the Earth


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.

hR + h

R


Variation of acceleration of gravity with depth

Consider a body at a depth d below the surface of the earth. Thus, r= R- d .

dR

r = R - d


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 isgd = GM’/ r2 . If we assume that the earth has uniform density ρ, thenM’ = (4/3) πr3 ρTherefore, gd = G ((4/3) πr3 ρ)/ r2 = (4/3)Gπrρ

Therefore, gd = (4/3)Gπρ(R-d)


Variation of g on the surface of the earthAlthough 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.


Thus, total force F, F= Weight of the object + Centripetal forceF= W + mRω2 (Where ω is the angular velocity of the earth)W= F - mRω2

mg= GMm/R2 – mRω2

g = g0 – Rω2

Axis of rotation

λ

At an inclination of λ from the equator (latitude λ) , it can be shown that,

g = g0 – Rω2Cos2λ

• 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


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.

g1 = GM1 / R12 g2 = GM2 / R2

2

g1 / g2 = (M1 / M2) (R2 / R1 )2

R1 R2

M1 M2


Gravitational Potential EnergyDefinition-Gravitational 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)


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.


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.


Gravitational potential at a pointIt 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).


Escape VelocityIt 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

Therefore, ve = √(2gR)Ve – Escape Velocity


Motion of planets and satellites

Consider a satellite of mass m moving around a planet of mass M in a circular orbit radius r. IfV0 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.


Time period of revolutionT = 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 = r1

3

T22 r2

3


Energy of a satelliteThe 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


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.


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 × ΔrForce 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 GradientBut, potential decreases when r increases.So, EG = - dV/dr


Equipotential SurfacesThe gravitational potential outside a spherical mass M at a point distance r from its center is given by V= -GM/rSo, 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.

Equipotential surfaces

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


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Gravitation