GRAVITATIONAL FIELD

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

M₁

M₂

The law can be expressed mathematically as:

F ά M₁M₂………………… 1

F ά 1

r²……………………2

F ά M₁M₂……………………3

r²

:. F = G M₁M₂……………………4

r²

M₁ and M₂ are the masses of the two particles r is the distance between them and G is the universal gravitational constant. The numerical value of G = 6.67 x 10^-11 Nm² 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

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

R²

But g = F …………………2

:. g = GMm x 1

r² m

g= GM …………3

r²

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

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

M_e = mass of the earth, m = mass of the satellite

Then F = GM_em

r₂

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

r²

This work must equal the Kinetic energy of the body of mass m at this point, having a velocity, V_e

Thus KE = ½ mV²

:. ¹/₂mV_e² = GM_em x r

r²

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

Then V_e² = 2GM_e

V_e² = 2Gm x R

R²

But g = Gm

R²

Ve² = 2gR

Ve² = √2gR

Energy in Gravitational Field

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

The centripetal force = mv² = GMm

r² r²

KE = ½ mv² = GMm

PE of mass in orbit = – GMm

The total energy in orbit = PE + KE

= – GMm + GMm

r 2r

= GMm

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 ^-11Nm²kg^-2
Calculate the escape velocity of a satellite from the earth’s gravitational field (g = 9.8m/s2, R = 6.4 x 10 ⁶m)
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 = 6.4×10⁶}
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 10⁶m (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/s². G = 6.7 x 10-11Nm2kg-2, calculate the mass of the earth if the radius is 64000km (a) 6.14 x 10²³kg (b) 5.99 x 10²⁴kg (c ) 3.98 x 10²⁶kg (d ) 4.02 x 10²⁵kg

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^-27kg and the mass of an electron is 9.11×10^-31kg, calculate the force of gravitation between: (i) a proton and an electron (ii) two electrons (iii0 two protons [G= 6.67×10^-11Nm²kg^-2; distance between the protons = 4.0m; distance between the electrons = 2×10^-2m; distance between the proton and the electron = 5.4×10^-11m]
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 NM²Kg^-2 )
If the mass of the earth is 5.78 x 10 ²⁴Kg and gravitational constant is 6.67 x 10^-11 Nm²kg^-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^-27kg and the mass of an electron is 9.11×10^-31kg, calculate the force of gravitation between: (i) a proton and an electron (ii) two electrons (iii) two protons {Take G= 6.67×10^-11Nm²kg^-2, distance between the protons = 4.0m., distance between the electrons = 2×10^-2m, distance between the proton and the electron = 5.4×10^-11m}