About Gravitational field

1. SS2
www.alhijrahcollege.comPage 1
Gravitational Field
This is any region or space around a mass in which the gravitational force of the mass is felt. The gravitational
force is a force field which acts when two bodies are not in contact.
Gravitational force between two masses
Gravitational force exists between two objects on the earth’s surface at a distance from each other. The pull
of gravity acts on mass of all sizes.
It is a vector quantity.
Newton’s Law of Gravitation
This law states that, the force of attraction between two bodies is directly proportional to the product of their
masses and inversely proportional to the square of their separation (i.e their distance apart).
Mathematically,
Fα mM α 1/r2
F = GmM/r2; where mM are the masses, r, is their distance apart, F is the force and G is the gravitational
constant whose value is 6.67 x 10-11 Nm2 kg-2.
Relationship between Acceleration due to gravity, g and the Gravitational constant, G
The earth is a sphere of radius, R, with mass, M concentrated at its centre. The distance of any object on the
earth’s surface to the centre of the earth is R, the earth’s radius. The gravitational force of attraction of the
earth on any mass, m, on the earth’s surface is given by:
F = GmM/R2.
This is the force of gravity on the mass due to the earth, i.e the weight of the object, mg, where g is the
acceleration due to gravity.
Thus,
F = mg = GmM/R2 ----------------- (equation *)
Hence the force per unit mass F/m = g, which means that g can be considered as the gravitational field
strength. Its S.I units are (m/s2) and (N/kg).
It is approximately constant at 9.8 Nkg-1 near the surface of the earth.
From equation * above,
g = GM/R2.
This is the gravitational field strength due to the earth. Its value decreases with increasing distance fromits
centre of mass.

2. SS2
www.alhijrahcollege.comPage 2
Acceleration due to Gravity on different Planets
For an object that is thrown upwards, the maximum height it reaches depends on:
i) the initial velocity, u
ii) the acceleration due to gravity, g
If we consider an object which is thrown vertically upwards with an initial velocity, u,
v2 = u2 - 2gh
Now, at the instance the object reaches its maximum height, v = 0,
Thus, we have
u2 = 2gh.
If the objects are thrown upwards on different planets, but each time with the same initial speed, we can say,
u2/2 = gh
i.e gh = a constant. (u2/2)
In other words, the height the object will reach is inversely proportional to the acceleration due to gravity. i.e
g α 1/h
So, when comparing different planets
g1h1 = g2h2
Therefore, g1/ h1 = g2/ h2
Worked example
A person can jump 1.5 m on the earth. How high could the person jump on a planet having twice the mass of
the earth and twice the radius of the earth?
Solution
Let mass of the earth = M; let the radius of the earth = R
Mass of the other planet = 2M; radius of the other planet = 2R
Let the acceleration due to gravity on the earth and the other planet be g1 and g2 respectively.
Height h1 reached by the person on earth = 1.5 m
Let the height reached by the person on the other planet = h2
Now, since g = GM/R2,

3. SS2
www.alhijrahcollege.comPage 3
Therefore
for earth, g1 = GM/R2
for the other planet, g2 = G.2M/(2R)2 = GM/2R2
since g1/ h1 = g2/ h2, therefore g1/ g2 = h2/ h1
g1/ g2 = (GM/R2)/(GM/2R2) = 2
therefore, 2 = h2/ h1 , 2 = h2/ 1.5
h2 = 2 x 1.5 = 3 m
Weight of Objects in Space
When an object is on or near the surface of the earth, it experiences the full extent of the force of gravity or
acceleration due to gravity of the earth. The gravitational force of attraction between the object and the earth
is given as:
Case 1: When the object is on the surface of the earth
F = GMm/R2
Case 2: When the object is above the surface of the earth
F = GMm/(R+h)2
Case 3: When the objects is below the surface of the earth
F = GMm/(R-h)2
mM
R
R
m
h
M
R
M
m
R

4. SS2
www.alhijrahcollege.comPage 4
Worked Example
Calculate the force of attraction between the earth and an object of mass 20 kg if
(i) the object is on the surface of the earth
(ii) distance 30000m above the surface of the earth
(iii) distance 30000m below the surface of the earth.
[R = 6.4 x 106m, M = 5.97 x 1024 kg, G = 6.7 x 10-11 Nm2kg-2]
Solution
m = 20 kg, R = 6.4 x 106m, M = 5.97 x 1024 kg , F = ?
(i) F = GMm/R2 = (6.7 x 10-11 x 5.97 x 1024 x 20)/( 6.4 x 106)2 = 195.3 N
(ii) F = GMm/(R+h)2 = (6.7 x 10-11 x 5.97 x 1024 x 20)/( 6.4 x 106 + 30000)2 = 193.5 N
(iii) F = GMm/(R-h)2 = (6.7 x 10-11 x 5.97 x 1024 x 20)/( 6.4 x 106 - 30000)2 = 197.2 N
Gravitational Potential
The gravitational potential at a point in a gravitational field is defined as the workdone by an external agent in
bringing a unit mass from infinity to that point. It is given as Vgrav = -GM/r (Jkg-1). Where M is the mass
producing the gravitational field, r , is the distance apart and G is the gravitational constant.
Gravitational potential, Vgrav, is the potential due to the gravitational field of the earth (or planet). Thus, it is
also defined as the gravitational potential energy, (Egrav) given to an object per unit mass:
Vgrav = - Egrav/m
= - [GMm/R] / m
= - GM / R
Gravitational Potential Energy
Gravitational potential energy, Egrav, at a point in a gravitational field is defined as the workdone in bringing a
mass, m, from infinity to that point.
It is given as Egrav = GMm/r = m x Vgrav (J).
For short changes in distance Δh which is far less than R (Δh<<R), gravitational potential energy is given by:
Egrav ≈ m g Δh
Worked Example
What is the gravitational potential at the Earth’s surface?

5. SS2
www.alhijrahcollege.comPage 5
[Mass of the Earth = 5.97 x 1024kg, Earth’s radius =6371 km, G = 6.7 x 10-11 Nm2kg-2]
Solution
Mass = 5.97 x 1024kg, R = 6.371 x 106m, G = 6.7 x 10-11 Nm2kg-2, Vgrav = ?
Vgrav = - GM / R
= - (6.7 x 10-11 x 5.97 x 1024) / 6.371 x 106
= - 6.278 x 107 J/kg
Escape Velocity
This is the minimum speed needed for an object to “break free” from the gravitational attraction of a massive
body such as the earth. That from the earth is about 40,270 km/h (11186.11 m/s).
It is also the velocity at which the sum of an object’s kinetic energy and its gravitational potential energy equal
to zero.
That is:
Gravitational Potential Energy + Kinetic Energy = 0
Therefore,
- [GMm/R] + ½ mv2 = 0
½ mv2 = GMm/R
v2 =2GM/R
v = square root [2GM/R] ie
The escape velocity can also be expressed in term of g.
Since g =GM/R2
Note:
1. At escape velocity, the object will move away forever from the massive body (such as earth), without
additional acceleration applied to the object.
2. As the object moves away from the massive body, the object will continually slow and (asymptotically)
approach zero speed as the distance of the object approaches infinity.
Worked example
What is the escape velocity of a body launched from the earth’s surface? [g = 10 m/s2, R = 6.4 x 106 m]

6. SS2
www.alhijrahcollege.comPage 6
Solution.
g = 10 m/s2, R = 6.4 x 106 m, Escape velocity, v = ?
v = squareroot [2 g R] i.e
v = (2 x 10 x 6.4 x 106)0.5 = 1.13 x 104 m/s.
Satellites
Bodies that orbit the larger and more massive bodies are regarded as satellites. The moon orbits round the
Earth, Phobos and Deimos revolve round the Mars, these are satellites.
Similarly, the Earth, Mars and other planets that orbit the Sun are regarded as the Sun’s satellites.
Natural and Artificial Satellites
The natural satellite is an orbiting body that occurs naturally in space. Eg moons that orbit planets are natural
satellites.
The artificial satellite is a man-made satellite that is made to orbit the planet (earth). Most artificial satellites
revolve around the Earth. Their functions include the monitoring of the global weather, telecommunications,
military and intelligence surveillance. Example of such is Hubble Space Telescope.