Physics

By the end of the unit, learner should be able to explain gravitational field potential and its application in planet motion.

My goals

By the end of this unit, I will be able to:

* explain universal gravitation field.

* describe the factors affecting force of gravity.

* state and explain Kepler’s laws of planetary motion.

* investigate planetory motion using computer simulation.

Link to other subjects

Geography and Astronomy (Landslides, motion of planets and satellites) Chemistry (Electrons orbiting the nucleus).

Introduction

The Universe is composed of different planets one of which is the earth.

All objects on the earth remain on it. They cannot move away unless acted on by external forces. This shows that there is a region around it that provides a force that attracts these earthly objects.

Since the earth is part of the universe it follows that a round the universe there is attracting field.This is called universal gravitational field.

Universal gravitational field potential

To have potential is to have energy, therefore gravitational field potential is the ability of gravity to attract other objects.

1. What force that unites us as Banyarwanda?

2. How do you feel if you come close to a fellow munyarwanda when you find him/her outside our country? Relate the situation to the force around the earth.

3. What makes you feel attracted to your fellow munyarwanda?

A field is a region of space where forces are exerted on objects with certain properties.

The diagram represents the Earth’s gravitational field. The lines show the direction of the force that acts on a mass that is within the field.

This diagram shows that:

• Gravitational forces are always attractive – the Earth cannot repel any objects.

• The Earth’s gravitational pull acts towards the centre of the Earth.

• The Earth’s gravitational field is radial; the field lines become less concentrated with increasing distance from the Earth.

The force exerted on an object in a gravitational field depends on its position.The less concentrated the field lines, the smaller the force. If the gravitational field strength at any point is known, then the size of the force can be calculated.

The gravitational field strength g at any point in a gravitational field is the force per unit mass at that point:

Close to the Earth’s surface, g has the value of 9.81Nkg^-1, though the value of 10Nkg^-1 is often used in calculations.

Gravitational field strength is a vector quantity: its direction is towards the object that causes the field.

In studying gravitation, Newton concluded that the gravitational attractive force that exists between any two masses:

• Is proportional to each of the masses.

• Is inversely proportional to the square of their distances apart.

The law states that‘The force of attraction between two masses m₁ and m_2a distance r apart is directly proportional to the product of masses and inversely to the square of distance r of separation.’

This force acts along the line joining the two particles. In magnitude, the force is given by:

where m₁ and m₂ are the masses of the two particles, r is the distance between the centres of mass and G = 6.67 x 10^-11Nm² kg^–2 is the universal constant of gravitation.

Dimensions of gravity

The S.I unit of G is Nm² kg^–2

A point mass is one that has a radial field, like that of the Earth.

A graph that shows inverse square law

Gravitational potential energy

Potential and potential energy

Question about fig. 9.3

• The car at the top of the hill has more potential energy than the one at the bottom, but relative to ground level they both have zero. why?

• Note and record in your notebook your analysis.

Using this reference point

:• All objects at infinity have the same amount of potential energy; zero.

• Any object closer than infinity has a negative amount of potential energy, since it would need to acquire energy in order to reach infinity and have zero energy.

The gravitational potential at a point in a gravitational field is the potential energy per unit mass placed at that point, measured relative to infinity.

Calculating potential and potential energy

When an object is within the gravitational field of a planet, it has a negative amount of potential energy measured relative to infinity. The amount of potential energy depends on:

• The mass of the object.

• The mass of the planet.

• The distance between the centres of mass of the object and the planet.

The Centre of mass of a planet is normally taken to be at its centre.

The gravitational potential energy measured relative to infinity of a mass, m, placed within the gravitational field of a spherical mass M can be calculated using: 

Gravitational potential, V, is given by the relationship:

Gravitational potential is measured in Jkg ^–1.

Relation between the universal gravitational constant and force of gravity (g and G)

A small object of mass m, placed within the gravitational field of the Earth, mass M, experiences a force, F, given by:

Where r is the separation of the centres of mass of the object and the Earth.

It follows from the definition of gravitational field strength as the force per unit mass that the field strength at that point, g, is related to the mass of the Earth by the expression:

The same symbol, g, is used to represent:

• Gravitational field strength.

• Free-fall acceleration.

Kepler’s Laws

Activity1: Field work

As a class, let us visit one of the roundabouts (where three roads meet).Try to see/check how cars, motorcycles, bicycles move around it.

Qn i) Does the features on a roundabout move?Assuming a roundabout to be a sun and vehicles to be planets, what can you say?

1. Discuss your findings in groups of 5 members.

2. Present your findings to the whole class.

3. Note down the observation.

4. Present your work to the teacher for marking.

Activity 2

Check on the watch (that one with a clock hand).Look at where the second hand is fixed.While the hand is rotating about a fixed point,describe the shape the second hand describes!

Figure 9.4: Rotation about a fixed point

We can relate the movement of the minute hand as the movement of planets about the sun.

Kepler’s first law:The path of each planet about the sun is an ellipse with the sun at one focus(or planets describe ellipse about the sun as one focus).

Kepler’s second law: The line joining the sun to the moving planet sweeps out equal areas in equal times.

If planet P takes the same time to travel from 1 to 2 as from 3 to 4 then the shaded areas are equal.

Kepler’s third law: The squares of the times of revolution T of the planets about the sun are proportional to the cubes of their mean distances r from it: = = constant

The value of this constant is , M is the mass of the sun in this case.

Proof of Kepler’s third law

Activity 3

• Using Newton’s law of gravitation (Formula) and the formula that keeps the planet in circular paths (Formula for centripetal force), Derive expression for Kepler’s third law of planetary motion

• Put your derivation in your notebook after discussing it with your friends.

Planetary data applied to Kepler’s third law

EXAMPLES

1. Calculate the force of gravity between two bowling balls each having a mass of 8.0kg, when they are 0.50m apart.

2. At the surface of a certain planet, the gravitational acceleration g has a magnitude of 2.0m/s². A 4.0kg brass ball is transported to this planet. Give:

a) The mass of the brass ball on the earth and on the planet; and

b) The weight of the brass ball on the earth and on the planet

Exercises

1. Calculate the effective value of g, the acceleration of gravity,

(a) 3200m,

(b) 3200km, above the earth’s surface.

2. Determine the net force on the moon (m_m = 7.36×10²² kg) due to the gravitational attraction and both the earth (m_e=5.98×10²⁴ kg) and the sun (m_s=1.99×10³⁰ kg) assuming they are at right angles to each other.

3. What is the effective value of g at a height of 1000km above the earth’s surface? That is, what is the acceleration due to gravity of objects allowed to fall freely at this altitude?

4. The universal attraction is given by: FGd= where G = 67×10⁻¹² USI

a) Find the acceleration g_o at the earth’s surface in function of R, M and G, where R is the radius of the earth and M its mass. If R = 6400km, g_o = 9.81m/s, calculate M.

b) Find in function of g_o, R and h, the acceleration due to the gravity g at a certain height h.

c) If the satellite is at the height h,

(i) Find the speed in function of go, R and h.

(ii) Find its value if h = 36000km.5. Venus is at average distance of 1.08×10⁸ km from the sun. Estimate the length of the Venusian year using the fact that the earth is 1.49×10⁸ km.

6. The planet Mars of mass m describes around the sun of mass M, an ellipse of mean radius of orbit a = 230×10⁶ km in 1.8 years. The satellite Deimos of mass m’ describes around the planet mars an ellipse of mean radius a' = 28×10⁶ km in 30h. Find the mass of the planet mars, given that M = 2×10³⁰ km and 1year is 365days.

Extension Questions

1. We actually know fifteen satellites revolving around the planet Uranus. Let us denote the period of revolution of satellite by T and the mean distance to the centre of the planet by r. The five bigger than others have the following characteristics:

a) (i) For each satellite, calculate T₂ and r₃,

(ii) Assume T₂ = y and r₃ = x. Trace the graph of y = f(x). What conclusion related to the nature of the graph can you get?

b) (i) Calculate the slope of the plotted segment.

(ii) Deduce the mass of Uranus.