CHAPTER 7:

GRAVITATIONWONG LIHERNGTAN SI YOUNGTAN SZE MING

YEE SIEW QI

7.1 NEWTON’

SLAW

OFGRAVITAT

ION

Newton ‘s law of gravitation states that the force of attraction between particles is directly proportional to their mass and inversely proportional to the square of distance apart.

whereThe negative sign shows that the force is an attractive force.

Case 1: Determine the force of gravitational attraction between the earth 5.98 x 1024 kg and a 70 kg boy who is standing at sea level, a distance of 6.38 x 106 m from earth's center.  m1 = 5.98 x 1024 kg, m2 = 70 kg, r = 6.38 x 106 m, G = 6.6726 x 10-11N-m2/kg2

7.2 GRAVITATIONAL FIELD STRENGTH

• A gravitational field is a region where gravitational force acts on massive bodies. Eg . gravitaional field of the Earth.

• The gravitational field strength tells us how strong a gravitational field is. The gravitational field strength of the Earth near its surface is 9.81m / s2. This means an object that is near the surface of the earth will accelerate towards it at 9.81ms2. We could then define the gravitational field strength as the acceleration an object will experience within that gravitational field.

• A better definition, however, can be derived from the equation, F = ma.

• The gravitational field strength , E at a point is the force of gravity per unit mass exerted on a mass placed

• Gravitational field strength is a vector quantity which measured in ， though it is perfectly acceptable to use ms − 2 for situations where it is treated as an acceleration (such as the acceleration of an object in free fall).

• （ r > R ）

• The –ve sign shows that the direction of E is in the direction of decreasing r, that is, towards the centre of the Earth.

• (by substituting F for mg)

• (by cancelling the lower case 'm's)

• On the surface of the Earth ， r=R and

Questions:

• 1. A point mass produces a gravitational field strength of 3.4 x 10-2 N/kg at a distance of 4.9 x 102 m away. What is the mass of the point?

Ans:

g = Gm/r2 Thus, m = gr2/G = 1.2 x 1014 kg

• 2. Estimate the gravitational field strength at the surface of an interstellar body whose density is 5.5 x 103kg/m3 and radius is 6.4 x 106 m.

Ans : g = Gm/r2 and m = density x volume = D x V = D x 4/3 x p x r3 Substitute for m in the field equation above. Thus, g = 4pGDr/3 = 4 x p x 6.67 x 10-11 x 5.5 x 103 x 6.4 x

106/3 = 9.8 N/kg

Variation of Acceleration Due to Gravity g with Distance r from the centre of

the Earth

• Weight = mg = gravitational force• Acceleration due to gravity, g=gravitational force/m =gravitational field strength,E # The variation of the acceleration due to gravity g with distance r

from the centre of the Earth equals the variation of the gravitational field strength.

• For a mass m on the surface of the Earth, weight =mg =

Acceleration due to gravity on the surface of the Earth ,

Where g = accleration due to gravity G = universal gravitational constant M = mass of the earth R = radius of the earth

• The variation of the acceleration due to gravity g’ with distance r from the centre of the Earth is illustrated by the graph above.

• Figure above shows a mass m at a distance of r < R from the centre of the Earth.

Variation of g with latitude• The acceleration due to gravity, and the gravitational field strength on the

surface of the Earth is not the same at different points on the Earth’s surface because :-

• The Earth is not a perfect sphere, but it is an ellipsoid as shown in below.• The rotation of the Earth about its axis.

• The Earth’s rotation on its axis causes it to bulge at the equator and flatten at the poles, forming an ellipsoidal shape.

• The figure below shows that the radius of the Earth at the equator R1 is greater than the distance of the poles from the centre of the Earth R2.

• g α ⅟r²• The value of g at the equator is less than the value of

g at the poles.

-The centripetal force m(R cos Ѳ)ω² is providedby the component of moIn the direction to the axisof the rotation.-At the N-pole, Ѳ=90° g’=g-At the equator, Ѳ=0° g’=g-Rω²

7.3 GRAVITATIONAL

POTENTIAL

Gravitational Potential

• The strength of the gravitational force at a point in a gravitational field is described by the gravitational field strength E or g is a vector quantity.

• Another quantity associated with the point in the gravitational field is the gravitational potential. It is a scalar quantity.

• The gravitational potential V at a point P in a gravitational field is defined as the work done per unit mass to bring a body from infinity to P. The unit for gravitational potential is Jkgˉ¹.

• The gravitational potential energy U of a body at a point P in a gravitational field is defined as the work done to bring the body from infinity to P. The unit for gravitational potential energy is J.

• Hence the gravitational potential energy U of a body of mass m at a point where the gravitational potential, V is given by

U = mV

Conversely,

U

V = m

• Figure below shows a body of mass m at a distance of x from the centre of the Earth.

The gravitational force on the body is

• The work done to move the body a small distance dx towards the Earth, that is, a displacement ( dx ) is

dU = F (-dx)

= (-G )( -dx)

Mm x²F = - G

PMR

rx

m-dx

F = - GMmx²

Mm x²

G Mm

x² dx=

• The work done to bring the body from infinity to the point P which is at a distance r from the centre of the Earth is

Gravitational potential energy, U = - G

• Gravitational potential at a distance r from the centre of the Earth is

V =

Gravitational potential, V = -• The negative sign in the expression for U and V shows that the

work is done by the Earth to bring the mass from infinity.

U

m

r

Mm

r

GM r

∞

• On the surface on the Earth, r = R– Gravitational potential, V = -

– Gravitational potential energy, U = -

• The graph illustrates the variation of the gravitational potential V with distance r from the centre of the Earth.

E α

V α

V

r

GM R

1r

0R

-

1 r²

GM R

GMm R

Changes in Gravitational Potential Energy• A body of mass m on the surface of the Earth has

gravitational potential energy, U = - G

• When the body is raised through a small height dR, the change in its gravitational potential energy dU is obtained by differentiating U with respect to R.

dU = G dR

= mg (dR)

• If dR is written as h, then the change in gravitational potential energy of a body of mass m when it raised through a height h is mgh.

∆ U = mgh

Mm R²

Mm R

7.4RELATIONSHIP

BETWEEN g AND G

G is the Universal Gravitational Constant.It is a scalar quantity with dimension

g is the acceleration due to gravity .It is a vector quantity with dimension

Where g = accerelation due to gravity

R = constant radius of earth

G = universal gravitational constant

M = mass of Earth

EXAMPLE :A planet of mass M and radius r rotates about its axis with an angular velocity large enough to substances on its equator just able to stay on its surface . Find in terms of M , r and G the period of rotation of the planet.

7.5 Satellite Motion in Circular Orbits

Satellite is a body that revolves round a planet. Satellites can be categorized as natural satellites or man-made satellites. The moon, the planets and comets are examples of natural satellites. Examples of man-made satellites are Sputnik I , Measat I ,II and III which are communication satellites.

In order to launch satellite into orbit , rockets are used. When rocket that carries the satellite reaches the required height , the satellite is launched into circular orbit with a certain velocity v that is tangential to intended orbit.

Since the centripetal force required for the circular motion of the satellite is provide by the gravitational attraction of the Earth on the satellite,

(mv2) / r = G(Mm) / r2

Velocity of satellite, v = [ (GM) / r]1/2

since GM= gR2, v = [ (gR2) / r]1/2

For a satellite orbiting close to the Earth’s surface , the radius of the satellite’s orbit is approximate to the radius of the surface.

( r = R)V = 7.92 x 103 ms-1

Period of satellite’s orbital , T = 85 minutes

Synchronous Satellite

Synchronous Satellite is the communication satellite in which the orbit of it is synchronised with the rotation of the Earth about its axis .It should have the characteristics of :Period of its orbit = 1.0 day (which is equals to the period of the earth’s rotation about its axis)

The satellite orbits the Earth in the same direction as the direction of the Earth.The satellite’s orbit is in the same plane as the equator because the centripetal force is provided by the gravitational attraction of the Earth.

As Qravitational force = Centripetal force G[ (Mm) / r2] = mrw2

r3 = GM [ T / (2π) ] 2

*Radius of satellite’s orbit , r = 4.24 x 10 7 m (6.6 times the radius of the Earth)

Planetary System

1. Planet is a body in orbit round the Sun which sastifies the condition of •Has enough mass to form itself into a spherical shape•Has cleared its immediate neighbourhood of all smaller objects

2. Due to the above definition of ‘planet”, Pluto was excluded from planethood and reclassified as a ‘dwarf planet”.

3.Planets orbit the Sun in elliptical orbits. As an

approximation, we assume that the orbits of the planets are circular.

If a planet of mass m is in a circular orbit of radius r round the Sun (mass Ms), the centripetal force is provided by the gravitational attraction between the planet and the Sun.

G[ (Mm) / r2] = mrw2

r3 = GM [ T / (2π) ] 2

T2/ r3 = (4 π2) / GMs = constant

(which sastifies Kepler’s Third Law)

Energy of a SatelliteTotal energy of the satellite ,

E = U+ K E = -(GMm)/r + (GMm)/2rAs an object approaches the Earth ,

*Its kinectic energy K increases as the gravitational pull of the Earth on it increase.

*Its gravitational potential energy U decrease , becoming more and more negative.*Its total energy E also increase.

Weightlessness

1.The apparent loss of weght is known as weightlessness.For example , an astronaut in a spacecraft feels that he has no mass as all objects in the spacecraft experience an apparent loss of weght.

2. A spacecraft orbiting the Earth has an acceleration that equals the

acceleration due to gravity g at the position of th spacecraft.

3. If N = normal reaction of the floorboard of the spacecraft on an object of mass m ,

Using F = ma,Mg – N = ma

N = 0*means that the floorboard does not exert

any force on the object.

4. Condition for a person to experience weightlessness :

The person must fall towards the Earth with an acceleration that equals the acceleration due to gravity.

7.6 Escape

Velocity

When a rocket is launched from the earth’s surface , it can escape from gravitational pull of the earth if it has sufficient energy to travel to infinity.

The energy required come from the kinetic energy when the rocket is launched .

The minimum velocity is also known as escape velocity .

Expression for escape velocityUsing the principle of conservation of energy,

Kinetic energy required by the body on the Earth’s surface

= Gain in gravitational potential energy of the body when it moves from the Earth ‘s surface to infinity.

Case 1: What is the escape velocity on earth with respect to Earth's gravity? Mass of the earth = 5.98x1024kg,

R = 6378100 m,

G = 6.6726 x 10-11N-m2/kg2.