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# Classwork Series and Exercises {Physics- SS3}: Potential Energy in Gravitational Field

Concept of Gravitational Field

There are two types of forces – contact forces and force fields. Contact forces are most common in everyday life. For example, you push or pull on wheelbarrow, a tennis racket exerts a force on a tennis ball when they make contact, your foot exerts a force on a football when you kick it. Force fields, e.g. gravitational forces, act even when the two bodies are not in contact. The earth, for example exerts a force on a falling mango fruit. It also exerts a force on the moon which is about 385,000km away. The sun itself exerts a force on the earth even though the earth is about 1.5 x 108 km distant from the sun.

In order to explain the observation of forces acting at a distance, it has been postulated that a gravitational field surrounds everybody that has mass, and this field fills up all of space. A second body at a particular location near the first body experiences a force because of the gravitational field that exists there. The gravitational field of the first body acts directly on the second body.

Gravitational field is a region or space around a mass in which the gravitational force of the mass can be felt.

Gravitational Forces between Two Masses

Gravitation is the force of attraction is the force of attraction exerted by a body on all other bodies in the universe. Hence a gravitational force exists between a body and all other bodies around it. Gravitational forces act between all masses and hold together planets, stars and galaxies, each mass has a gravitational field around it.

It was Sir Isaac Newton who first proposed the relationship between the gravitational force F, between two masses, m1 m2 and the distance, r, between these masses. He proposed his famous Law of Universal gravitation which we can state as follows:

Every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the centre of the two particles.

F µ m1 m2/r2

The magnitude of this force can be written as

F = Gm1m2/r2

Where m1 and m2 are the masses of the two particles, r is the distance between them and G is a universal constant of gravitation or simply gravitational constant.

Gravitational constant, G, has the same numerical value for all objects (G = 6.67 x 10-11 Nm2kg-2).

Newton’s law of gravitation refers to the force between two particles or bodies. Actually such a gravitational force is a pair of forces, an action-reaction pair. Although the masses of the particles may be different, forces of equal magnitude act on each other, and the action line of both forces lies along the line joining the bodies. Mass m1 attracts m2 with a force given by the equation F = Gm1m2/r2. By Newton’s third law of motion, action and reaction are equal and opposite. Therefore, m2 also attracts m1 by an equal but opposite force.

Gravitational attraction keeps the moon in its orbit around the earth and the earth in its orbit around the sun. Gravitational forces are always those of attraction.

Example

Two 5.0 kg spherical balls are placed so that their centres are 50.0 cm apart. What is the magnitude of the gravitational force between the two balls? (G = 6.67 x 10-11 Nm2kg-2)

Solution

F = Gm1m2/r2

= 6.67 x 10-11 x 5.0 x 5.0/0.52

= 6.67 x 25 x 10-11/0.52+

= 6.67 x 10-9 N

Because of the universal law of gravitation, there is a gravitational force of attraction between the sun and the planets, between earth and the moon, and also between other planets.

Relation between the Gravitational Constant ‘G’ and the acceleration of gravity at the earth surface ‘g’

The earth is supposed to be a sphere of radius, re, with its mass me concentrated at the earth’s centre. The distance of any object on the earth’s surface to the centre of the earth is re the earth’s radius. The gravitational force of attraction of the earth on any mass, m, on the earth’s surface is given by

F =Gmem/re2

This is the force of gravity on the mass due to the earth, that is, the weight of the object, mg, where ‘g’ is the acceleration due to gravity.

Thus, F =Gmem/re2 = mg

The force per unit mass, F/m, is given by

F/m = Gems/re2 = g

Hence g = Gme/re2

This means that the acceleration due to gravity ‘g’ can be considered as the force per unit mass on the earth’s surface. According to this equation g = Gme/re2, the acceleration of gravity, ‘g’ at the surface of the earth, is determined by me (the earth’s mass), and re (the earth’s radius), hence from equation g = Gme/re2 we should ‘g’ to be slightly greater at the top of the mountain   than at the sea level. This is what actually obtains in practice.

If ‘g’ and ‘G’ are actually known we can use equation g = Gme/re2 to calculate the earth’s mass, me.

Me = gre2/G.

The radius of the moon is one-fourth, and its mass is one eighty-first that of the earth. If the acceleration due to gravity on the surface of the earth is 9.8 ms-2, what is its value on the moon’s surface.

The relation between g and G is given by

g = Gme/re2

For the earth, ge = Gme/re2                            (i)

For the moon, gm = Gmm/rm2                       (ii)

Since G is the universal constant of gravitation it is constant for both equations.

r= re/4                                                                    (iii)

mm = me/81                                                               (iv)

From equation (i)

G = gere2/me                                                               (v)

Putting (v) in (ii) we have

gm =  gere2/me . mm/rm2

= 9.8 x (re/ rm)2 x mm/me                                  (vi)

From (iii) re/rm = 4

From (iv) mm/me = 1/81

Putting these in equation (vi) we have: gm = 9.8 x (4)2 x 1/81

= 9.8 x 16 x 1/81 = 1.9 ms-2

Gravitational Potential

The work done in raising a mass, m from the ground surface to a height, h above the ground is given by

W = mgh

The work has been done against the gravitational pull of the earth. This work appears as the gravitational potential energy (PE) of the body.

This potential energy is dependent of the height, h, or the relative position of the body from the ground or zero level where the PE is considered to be zero.

In general points in any gravitational field possess gravitational potential. If free to move, a body will tend to move from a point of higher gravitational potential to points of lower gravitational potential

Gravitational Potential (G) at a point is defined as the work done in taking unit mass from infinity to that point. Unit is jkg-1.

This gravitational potential is given by

V = -Gm/r

Where m is the mass producing the gravitational field and r is the distance of the point to the mass. The gravitational potential decreases as r increases and become zero where r is infinitely large, the negative sign in the equation above indicates that the potential at infinity (zero) is higher that the potential close to the mass.

Escape Velocity

There are many man-made satellites that circle around the earth at the present time. One common feature of these bodies is that they are held in an approximately circular path by the earth’s gravitational pull. It is this force that provides the needed earth’s centripetal force required to keep the satellites in their orbits. The velocity (vs) of the satellite as it orbits round the earth is given by

Mvs2/re = F = Gmme/re2 (centripetal force) = (Gravitational force)

Hence vs = ÖGme/re

This is the velocity with which the satellite moves round the earth. Notice that the mass of the satellite does not enter into this vs = ÖGme/re. All satellites in orbit with radius re must have the same speed. For a satellite to escape from the earth and never return, it must be launched with a velocity greater than that required to make it orbit.

We define the escape velocity (ve) as the minimum velocity required for an object (e.g. satellite or rocket) to just escape or leave the gravitational influence or field of an astronomical body (e.g. the earth) permanently.

We can obtain the formula for the Escape velocity using the Newton’s law of universal gravitation which is an inverse square law:

F = Gmem/r2

Let m be the mass of the satellite, and me, the mass of the earth.

The work done in carrying a mass m from a point at a distance r from the centre of the earth, to a distance so great that the gravitational field is negligibly weak is given by

W = F x r

But from equation 2.2, F = Gmem/r2

Hence, W = Gmem/r2.r = Gmem/r

This work must equal the kinetic energy of the body of mass m at this point, having a velocity ve. This kinetic energy is given by:

KE = ½ mve2

½ mve2 = Gmem/r

ve2 = 2Gme/r

ve = Ö2Gme/r

If we launch the mass m from the earth’s surface, where r = R, we then have that ve = Ö2Gme/R

But from equation me = gre2/G, me = gr2/G

Hence ve = Ö2G/R.gR2/G

ve = Ö2gR

Where R is the earth’s radius

Test and Exercise

1. Determine the mass of the earth if the radius of the earth is approximately 6.38 x 106 m, G = 6.67 x 10-11 Nm2kg-2 and g = 9.80 ms-2.

A. 5 x 1024 kg B. 5.98 x 1025 C. 5.98 x 1024 D. 5.6 x 1025 . Answer is C. 5.98 x 1024

2. Determine the force of attraction between the sun (ms = 1.99 x 1030 kg) and the earth (me = 5.98 x 1024 kg). Assume the sun is 1.50 x 108 km from the earth.

A. 3.53 x 1018  B. 3.53 x 1022 N  C. 3.05 x 1032 D. 3.3 x 1022  Answer is B. 3.53 x 1022

3. Which of the following is correct for Escape Velocity?

A. ve = Ö2Gr B. ve = Ö2Gr C. ve  = Ö2gr D. ve = Ö2g. Answer is  D. ve = Ö2g

4. The numerical value for gravitational constant G is

A. 6.42 x 10-11 Nm2 kgB. 6.67 x 10-11 Nm2 kg2  C. 7.8 x 10-11 Nm2 kg2  D. 6.67 x 10-15 Nm2 kg. Answer is B. 6.67 x 10-11

5. The expression F= Gm1m2/r2 indicates

A. The gravitational force of attraction B. The gravitational force of repulsion C. Gravitational force of constant acceleration D. Force of repulsion. Answer is A. The gravitational force of attraction