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  Gravitation and Extended Bodies Objects, Physics tutorial

Gravitational Potential Energy:

Gravitational force is the central force and relies only on distance of influenced object from force center. As it is the conservative force it can be derived from the potential energy function. Potential energy of the system of two point masses interacting with each other through gravitational force is

U(r) = - (GmM)r.

Let potential energy at infinity be zero. Now, potential energy is stated as

U(r) - U(∞) = ∞∫r F→(r→)dr

Where we have chosen one point to be at infinity. The force points from the location of mass m to the origin (location of M), and we also chose the path to go directly along a radial direction so that the magnitude

r^.dr→1 = dr1

Therefore we get from

F→ = (GmM/r2)dr^

that

U(r) - U(∞) = ∞∫r(-GmM/r2)dr1

(-GmM/r1)|∞r

But U(∞) = 0

Therefore U(r) = -GmM/r

Escape Speed:

Escape velocity is stated as the minimum velocity with which the body must be projected so that it overcomes gravitational pull of earth.

To project the object (satellite) and land it say, on moon, it could first be projected to land on the orbit whose period of revolution is similar as time taken for earth to rotate about its axis that is 24 hours (this orbit is referred to as parking orbit for satellite) with the speed of 8kms-1 and then later firing rocket again to reach escape speed in suitable direction to land on moon. To get escape speed we use the given analysis.

We know that certain amount of energy is needed to escape from earth. Escape speed will be determined considering fact that potential energy gained by satellite will be equal to kinetic energy lost if we neglect air resistance.

Let m be mass of escaping body and M the mass of the earth. The force F exerted on the body by the earth when the distance separating them is x from the earth's center is given by

F = GMm/x2

Work done, δW by gravity when body moves the distance dx upwards

is

δW = -Fdx = -GMm/x2dx

Negative sign illustrates that force acts in opposite direction to displacement therefore,.

Total work done while body escapes =

r∫∞(-GMm/x2)dx

Where r = radius of earth

Therefore total work done = -GMm[-1/x]r∞ = GMm[1/x]r∞ = GMm

If body leaves earth with speed v and just escapes from its gravitational field then, KE = Potential Energy.

i.e. 1/2mv2 = GMm/r

therefore v=√2GM/r

But g = GM/r2

Substituting, we get

v = √2gr

This equation provides expression for velocity of escape. Substituting values of g = 9.8ms-2 and r = 6.4 x 106 m escape speed is computed to be

V= 11kms-1

Variation of g with Height and Depth:

Let us suppose that g is acceleration because of gravity at distance a from centre of the earth where a >r. r is a radius of the earth. Then from the studies on weight:

g = GME/r2

Hence g' = GME/a2

Where ME is mass of the earth and G is universal gravitation constant.

Dividing

g'/g = r2/a2

or

g' = (r2/a2) g

Above the earth's surface, acceleration because of gravity g′ differs inversely as square of distance, a between object and center of the earth. In the same equation r and g are constants. g′ therefore decreases with height.

g' = [r2/(r + h)2]g = [1/(1 + (h/r)2)]g

g' = [(1 + h/r)-2g]

At height h above earth's surface, a = r + h

If h is very small compared to r (where r is 6400km) neglect powers of h/r higher than first

Therefore

g' = (1-2h/r)g

g - g' = reduction in acceleration due due to gravity

g' = (2h/r)g

If we suppose sphere to be of uniform density, then from knowledge of relation between g and G we have

g1 = GM1/b2 and g = GM/r2

Where M1 is a mass of sphere of radius b. The mass of the uniform sphere is proportional to its radius cubed, therefore

M1/M = b3/r3

But

g1/g = (M1/M)r2/b2

Therefore g1/g = b/r

or g1 = (b/r)g

Suppose earth has uniform density, acceleration because of gravity g is directly proportional to distance b from center. I.e., it decreases linearly with depth. At depth h below earth's surface, b = r - h

Therefore g1 =((r-h)/r)g = (1-h/r)g

But as density of earth is not constant, g, actually increases for all depths

Variation of g with Latitude:

Acceleration because of gravity has been observed to differ from location to location. This is because of the following:

i) Equatorial radius of earth exceeding its polar radius by approx 21km therefore making g greater at poles than at equator as, a body is far from center of earth here.

ii) The effect of earth's rotation.

Fundamental Forces in Nature:

Fundamental or basic forces are those for which we can't find the underlying force from which they are derived. It then stands to reason that those forces resultant from operation of some fundamental force are called as derived forces. There are three types of fundamental forces. These are (i) gravitational (ii) electroweak and (iii) strong. Gravitational force that acts on all matter differs inversely as square of distance but its range is infinite. This force is liable for holding together planets and stars and in fact, in general organization of solar system and galaxies.

Electro weak force comprises electro-magnetism and supposed weak nuclear force. Electromagnetic forces comprise force between two charged particles at relative rest (electrostatics) or in relative motion (electro-dynamics). Electrostatic between two charges follows inverse square law like gravitational force between two masses. Difference here is that charges are of two types - positive and negative. If charges are of opposite type force between them is attractive but if they are of similar kind, force is repulsive. It can be illustrated that gravitational force between electron and the portion in the hydrogen atom is 1039 times weaker than electrostatic force between them. Therefore we get the comparative estimate of strengths of gravitational and electrostatic force.

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