Gravitational Field

CONTENT

• Introduction
• Law of Universal Gravitation
• Gravitational Potential
• Escape Velocity
• Potential Energy in 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 exists between a body and all other bodies around it. 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 particles with a force that is proportional to the products of their masses and inversely proportional to the square of the distance between them

EVALUATION

1. State Newton’s law of universal gravitation.
   1. Two small objects of masses 100Kgand 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 )

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.

RELATIONSHIP BETWEEN g AND G

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

                                    F = Gm₁m₂…………………..(1)

                                                r²

                                    But g = F    ………………..  ( 2 )

                                                m₂

                                     g = (Gm₁m₂ )/ r²  x 1 / m₂

                                     g / r²=Gm₁ ………………. (3)

Note that m₁ is the mass of the body exerting the force on the other (e.g the Earth).

EVALUATION

1. If the mass of the earth is 5.78 x 10 ²⁴Kg and gravitational constant is 6.67 x 10^-11 Nm²kg². Calculate the gravitational field intensity due to earth.  Radius of earth is 6400km
2. The gravitational field intensity at a location X in space is two-fifth of its value on the earth surface. If the weight of an object at x is 4.8N, what is the weight on earth?

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.

EVALUATION

1. Calculate the gravitational potential at a point on the earth surface. Mass of earth is 6.0×10²⁴kg, radius of earth = 6400km and G = 6.67 x 10 ^-11Nm²kg^-2
2. Calculate the gravitational potential at a point on the earth surface [ radius of the earth = 6.4X10⁴m  mass of  the earth = 6.0 X 10 ²⁴Kg, G= 6.67X 10 ^-11 Nm²Kg^-2]

ESCAPE VELOCITY

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

If M = mass of the earth, m = mass of the satellite, then the gravitational force of the Earth on the satellite is:

The following conclusions can be drawn from the equation.

i. The magnitude of the total energy is equal to that of the k.e of the satellite.

ii.The kinetic energy of a satellite in an orbit increases as the radius o the orbit decreases.

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

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

EVALUATION

Derive an expression for the total energy in a gravitational field.  What conclusions can you draw from the equation?

READING ASSIGNMENT

New School Physics for Senior Secondary Schools (M.W Anyakoha Pages 370-377).

GENERAL EVALUATION

1. An object is projected with a velocity of 50m/s  from the ground level  at an angle of φ to the vertical. If the total time of flight is 5s, calculate the value of φ.
2. A ball is projected horizontally from a height of 30m above the ground with a speed of 20m/s. calculate the horizontal distance travelled by the ball when it hits the ground  g = 10m/s²]

WEEKEND ASSIGNMENT

1.         What is the gravitational potential due to a molecule of mass m at a distance r from it.?

            (a) Gm             (b) Gm                         (c) Gm²            (d) m²

r²                             r                                   r                                   Gr²

1. 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

3.         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

4.         Two objects of mass 80kg and 50kg are separated by a distance of 0.2m. calculate the gravitational attraction between them if G = 6.67 x 10-11 Nm²kg^-2.

            (a) 6.67 x 10-6             (b) 6.67x 106N             (c) 5.92 x 10-6N          (d) 5.92 x 106N.

5.         Which of the following statements is/are correct about gravitational potential?

V = -Gm

r

I.          The negative sign indicates that work done is against the gravitational field.

II.        The potential at infinity is zero

III.       The potential at infinity is less than that at the surface of the earth.

IV.        Gravitational potential is a vector quantity

(a)I and II only    (b) III and IV Only    (c) III only  (d) I,II and III only  (e) II and IV only.

THEORY

1.         The average radius of Jupiter’s orbit round the sun of mass 2 x 10³⁰Kg is 7.8 x 10¹¹m.  If the mass of Jupiter is 1.9 x 10 27, find the gravitational force the sun exerts on Jupiter , G = 6.67 x 10^-11Nm²kg^-2

2.         If the mass of a portion is 1.67 x 10^-27kg and the mass of an electron is 9.11 x 10^-31kg, calculate the force of gravitation between:

i.          a proton and an electronii.        two electronsiii.            two protons.

Take G= 6.67 x 10-11 Nm2kg-2, distance between the protons  = 4.0m., distance between the electrons = 2 x 10-2m, distance between the proton and the electron = 5.4 x 10-11m