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

> 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.

where G is the universal gravitational constant

G = 6.67 x 10^{-11} m^{3} 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

when the distance is 3r, the force is

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 10^{32} kg and 9.0 x 10^{30} kg are 1.4 x 10^{16} 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?

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,

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.

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.

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

Example 4

The mass of a star is 2.0 x 10^{30} kg and its radius is 1.5 x 10^{6} m.Find the gravitational field strength at:

(a) The surface

(b) 1.0 x 10^{6} m above the surface of the star [G = 6.67 x 10^{-11} N m^{2 }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 m^{2} 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.

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.

Unit: J and it is a scalar quantity.

> The relationship of V and U is

> The relationship of F and U is

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

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 10^{10} m from the center of a star of mass 8.45 x 10^{35} kg.[G = 6.67 x 10^{-11} N m^{2} 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 10^{7} m above the earth? Given that the mass of the earth is 6.00 x 10^{24} kg and its radius is 6.40 x 10^{6} m. [G = 6.67 x 10^{-11} N m^{2} 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.

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.

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

Therefore v^{2} = gR

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

· Kinetic energy,potential energy and total energy of satellite

Total energy = Potential energy + kinetic energy GMm 1 GMm

Example 8

A satellite orbits the earth with period of rotation 2 hours. Given that the mass of the earth 6.0 x 10^{24} kg and the radius of the earth is 6.4 x 10^{6} m. Calculate:

(a) The height where the satellite is orbiting.

(b) The speed of the satellite. [G = 6.67×10^{-11} N m^{2} 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 and the gravitational potential energy at infinity is 0.

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

> This energy is being supplied by the kinetic energy

> Therefore, the escape velocity

Example 9

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

[Mass of the sun = 2.0 x 10^{30} kg, radius of the sun = 7.0 x 10^{8} m, G = 6.67 x 10^{-11} N m^{2} 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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