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Chapter 7 Gravitation

Introduction

All objects with mass attract each other with a gravitational force. The magnitude of the force between any two objects depends on their masses and the distance between the two objects.

7.1Newton’s Law of Universal Gravitation

clip_image002

> Newton’s Law of universal gravitation states that the force of attraction between two objects is directly proportional to their masses and inversely proportional to the square of their distance apart.

clip_image004

clip_image006where G is the universal gravitational constant

G = 6.67 x 10-11 m3 kg-1 s-2

Note:

1) Gravitational forces are very weak unless it is for objects with enormous mass such as planets and stars.

2) Gravitational force is a mutual attraction. We attract the Earth with the same force that the Earth attracts us.

3) Newton’s Law of universal gravitation is an example of an inverse square law. The force decreases in proportional to the square of the distance.

So when the distance is 2r, the force is clip_image008

when the distance is 3r, the force is clip_image010

Example 1

Why is it that the gravitational force of earth causes the object to be accelerated to earth but not otherwise ?

Solution

 

 

 

Example 2

Two stars of mass 6.0 x 1032 kg and 9.0 x 1030 kg are 1.4 x 1016 m apart. The net gravitational force is zero at a point P between the two stars and the distance of this point from the bigger star is r. What is the distance r?

clip_image012

Solution

 

 

 

 

7.2 Gravitational Field Strength

A gravitational field represents a region where a body having a mass experiences gravitational force.

> Gravitational field strength is defined as the force of gravity per unit mass.

> Gravitational field strength, clip_image014

Unit: N kg-1 and it is a vector quantity.

Example 3

What is the gravitational field strength acting on an object with mass m placed on the surface of the earth? Taking the mass of earth as M and the radius of earth as R.

clip_image016

Solution

 

 

 

7.3 Relationship Between g and G

> A body of mass m at a place on the earth’s surface where the acceleration of free fall is experiences a force F = mg (i.e. its weight).

> Assuming the earth has mass M and radius R, we can also calculate the same force using, Newton’s law of gravitation.

So, clip_image018

Therefore, clip_image020

Then clip_image022

Variation of acceleration due to gravity g with height and depth on the earth

clip_image024

Example 4

The mass of a star is 2.0 x 1030 kg and its radius is 1.5 x 106 m.Find the gravitational field strength at:

(a) The surface

(b) 1.0 x 106 m above the surface of the star [G = 6.67 x 10-11 N m2 kg-2]

Solution

 

 

Example 5

What is the mass of the earth given that its radius is 6400 km and the gravitational field strength at the surface is 9.81 N kg-1. [G = 6.67 x 10-11 N m2 kg-2]

Solution

 

 

 

 

7.4 Gravitational Potential

> The gravitational potential V at a point in a gravitational field is defined as the work done in bringing a unit mass from infinity to that point. clip_image026

Unit: J kg-1 and it is a scalar quantity.

> The gravitational potential energy U of a mass m at a point in a gravitational field is defined as the work done in bringing the mass from infinity to that point. clip_image028

Unit: J and it is a scalar quantity.

> The relationship of V and U is clip_image030

> The relationship of F and U is clip_image032

> The relationship of g (or E) and V is clip_image034

Note: Gravitational potential is defined as V= 0 at infinity. Since gravitational potential increases as height increases, V must have negative values when it is lower than infinity.

Example 6

What is the gravitational potential V at 3.50 x 1010 m from the center of a star of mass 8.45 x 1035 kg.[G = 6.67 x 10-11 N m2 kg-2]

Solution

 

 

 

 

Example 7

What is the change in potential energy when a rocket of mass 40 000 kg moves from the surface of the earth to a height of 1.80 x 107 m above the earth? Given that the mass of the earth is 6.00 x 1024 kg and its radius is 6.40 x 106 m. [G = 6.67 x 10-11 N m2 kg-2]

Solution

 

 

 

7.5 Satellite Motion in Circular Orbits

Satellite

A satellite is an object orbiting around a larger body by gravitational attraction.

e.g. The moon is a natural satellite of earth.

The earth is a natural satellite of sun.

There are also artificial satellites launched for the purpose of monitoring the weather, space observation and global communications.

clip_image036

Energy of a satellite

• Velocity and period of satellite in a fixed orbit

To keep a satellite in circular orbit, the centripetal force needed is being supplied by the gravitational force.

So clip_image038

clip_image040

clip_image042 since GM=gR2

If the satellite is close to earth, then r » R

Therefore v2 = gR

Once v is found, the period of satellite T is given by clip_image044

· Kinetic energy,potential energy and total energy of satellite

Kinetic energy = clip_image046

= clip_image048

Kinetic energy = clip_image050

Potential energy = clip_image052

Total energy = Potential energy + kinetic energy GMm 1 GMm

= clip_image052[1]+clip_image050[1]

= clip_image055

Example 8

A satellite orbits the earth with period of rotation 2 hours. Given that the mass of the earth 6.0 x 1024 kg and the radius of the earth is 6.4 x 106 m. Calculate:

(a) The height where the satellite is orbiting.

(b) The speed of the satellite. [G = 6.67×10-11 N m2 kg-2]

Solution

7.6 Escape velocity

> Escape velocity is the velocity required by an object on the surface of the planet in order to escape from the planet to infinity. The object will not be pulled back by gravity.

> Take the earth for example, the gravitational potential energy on the surface of the earth is clip_image052[2] and the gravitational potential energy at infinity is 0.

So, the energy required to send an object to infinity is

0 – (clip_image052[3]) = clip_image057

> This energy is being supplied by the kinetic energy

So, clip_image046[1]= clip_image057[1]

> Therefore, the escape velocity clip_image059

clip_image061

Example 9

What is the escape velocity from the surface of the sun?

[Mass of the sun = 2.0 x 1030 kg, radius of the sun = 7.0 x 108 m, G = 6.67 x 10-11 N m2 kg-2]

solution

 

 

Example 10

what is the escape velocity from the surface of the earth?

g on the surface of the earth = 9.81 m s-2, radius of the earth = 6400 km]

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