Table of Contents

**INTRODUCTION**

Gravitational field is a region or space around a mass in which the gravitational force of the mass can be felt. Gravitation is the force of attraction exerted by a body on all other bodies in the universe. Gravitational force act between all masses and hold together planets, stars and galaxies. Each mass has a gravitational field around it.

**LAW OF UNIVERSAL GRAVITATION**

Newton’s law of universal of gravitation states that *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*

M_{1} |

M_{2} |

K

R

The law can be expressed mathematically as:

F ά M_{1}M_{2 }………………… 1

F ά __1__

r^{2 }……………………2

F ά __M _{1}M_{2}__

^{ }……………………3

r^{2}

:. F = G __M _{1}M_{2}__

^{ }……………………4

r^{2}

**M _{1}** and

**M**are the masses of the two particles

_{2}**r**is the distance between them and

**G**is the universal gravitational constant. The numerical

**value**of

**G**= 6.67 x 10

^{-11}Nm

^{2}kg

^{-2}.

**GRAVITATIONAL FIELD INTENSITY**

Gravitational field intensity at a point is the force per unit mass of an object placed at that point.

g = __F__

M

The unit is N/Kg. It is a vector quantity and it is regarded as acceleration due to gravity.

__Relation between g and G__

If the force of attraction (F) between two particles of matter separated by a distance r is given by:

F = __GMm__…………………..1

R^{2}

But g = __F __ …………………2

m

:. g = __GMm__ x__ 1__

r^{2} m

g= __GM__ …………3

r^{2}

This is the gravitational intensity

**GRAVITATIONAL POTENTIAL**

The gravitational potential at a point is the work done in taking a unit mass from infinity to that point. The unit is Jkg-1.

The gravitational potential, V, is given by

V = __Gm__

r

**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 becomes zero when r is infinitely large*. The negative sign indicates that the potential at infinity (zero) is higher than the potential close to the mass.

** **

**ESCAPE VELOCITY**

This is the minimum velocity required for an object (e.g. satellite, rocket) to just escape or leave the gravitational influence or filed of an astronomical body (e.g. the earth) permanently.

m_{e} |

r m

R

M_{e} = mass of the earth, m = mass of the satellite

Then F = __GM _{e}m__

r_{2}

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 is

W = __GMm __ x r

r^{2}

This work must equal the Kinetic energy of the body of mass m at this point, having a velocity, V_{e}

Thus KE = ½ mV^{2}

:. ^{1}/_{2}mV_{e}^{2} = __GM _{e}m__ x r

r^{2}

If the mass was launched from the earth surface where r = R.

Then V_{e}^{2} = __2GM _{e}__

R

V_{e}^{2} = __2Gm__ x R

R^{2}

But g = __ Gm__

R^{2}

Ve^{2} = 2gR

Ve^{2} = √2gR

__Energy in Gravitational Field__

__Energy in Gravitational Field__

A satellite moving in an orbit round the earth has both kinetic and potential energy

The centripetal force = __mv ^{2}__ =

__GMm__

r^{2} r^{2}

KE = ½ mv^{2} = __GMm__

r

PE of mass in orbit = – __GMm__

r

The total energy in orbit = PE + KE

= – __GMm __+ __GMm__

r 2r

= __GMm__

2r

The following conclusions can be drawn from the equation.

- The magnitude of the total energy is equal to that of the k.e of the satellite.
- The kinetic energy of a satellite in an orbit increases as the radius of the orbit decreases.

iii. The kinetic energy of a satellite in an orbit increases as the speed of the satellite increases.

- The potential energy of the satellite in orbit is twice its kinetic energy and of opposite sign.

** **

**ONLINE WORK**

- State Newton’s law of universal gravitation and give the mathematical relation
- Calculate the gravitational potential at a point on the Earth surface. Mass of earth is 6.0x 10
^{24}/kg, radius of earth = 6400km and G = 6.67×10^{-11}Nm^{2}kg^{-2} - Calculate the escape velocity of a satellite from the earth’s gravitational field (g = 9.8m/s2, R = 6.4 x 10
^{6}m) - What is escape velocity?

** **

**ASSIGNMENT**

**Section A**

- A satellite is in a parking orbit if its period is (a) equal to the period of the earth (b) less than the period of the earth (c) the square of the period of the earth (d) more than the period of the earth
- The magnitude of the gravitational attraction between the earth and a particle is 40N. if the mass of the particle is 4kg, calculate the magnitude of the gravitational field intensity of the earth on the particle (a) 10.0Nkg
^{-1}(b) 12.6Nkg^{-1}(c) 25.0Nkg^{-1}(d) 160.0Nkg^{-1} - What is the escape velocity of a satellite launched from the earth’s surface? {g=10ms
^{-2}; radius of the earth = 6.4×10^{6}} - Calculate the escape velocity for a rocket fired from the earth’s surface at a point where the acceleration due to gravity is 10m/s2 and the radius of the earth is 6.0 x 10
^{6}m (a) 7.8 x 103m/s (b) 1.1 x 104m/s (c ) 3.5 x 107m/s (d) 6.0 x 107m/s - If g = 9.8m/s
^{2}. G = 6.7 x 10-11Nm2kg-2, calculate the mass of the earth if the radius is 64000km (a) 6.14 x 10^{23}kg (b) 5.99 x 10^{24}kg (c ) 3.98 x 10^{26}kg (d ) 4.02 x 10^{25}kg

**SECTION B**

- (a) State Newton’s law of universal gravitation (b) write down the expression for the gravitational force between two masses. Explain the meaning of each term in your expression (c) the mass of proton is 1.67×10
^{-27}kg and the mass of an electron is 9.11×10^{-31}kg, calculate the force of gravitation between: (i) a proton and an electron (ii) two electrons (iii0 two protons [G= 6.67×10^{-11}Nm^{2}kg^{-2}; distance between the protons = 4.0m; distance between the electrons = 2×10^{-2}m; distance between the proton and the electron = 5.4×10^{-11}m] - Two small objects of masses 100kg and 90kg respectively are separated by a distance of 1.2m. Determine the force of attraction between the two objects. (G = 6.67 x 10
^{-11}NM^{2}Kg^{-2}) - If the mass of the earth is 5.78 x 10
^{24}Kg and gravitational constant is 6.67 x 10^{-11}Nm^{2}kg^{-2}. Calculate the gravitational field intensity due to earth. Radius of earth is 6400km - Derive an expression for the total energy in a gravitational field. What conclusions can you draw from the equation?
- If the mass of a proton is 1.67×10
^{-27}kg and the mass of an electron is 9.11×10^{-31}kg, calculate the force of gravitation between: (i) a proton and an electron (ii) two electrons (iii) two protons {Take G= 6.67×10^{-11}Nm^{2}kg^{-2}, distance between the protons = 4.0m., distance between the electrons = 2×10^{-2}m, distance between the proton and the electron = 5.4×10^{-11}m}

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