1. HSC Topic 9.2 Space 9.2.1 Gravitational force and field
9.2.1 Objectives
Define weight as the force on an object due to a gravitational field.
State Newton’s Universal Law of Gravitation.
Define gravitational field strength.
Determine the gravitational field due to one or more point masses.
Explain that a change in gravitational 
potential energy is related to work done.
Solve problems involving gravitational forces and fields.
© 2006 By Timothy K. Lund

2. Topic 9.2: Space 9.2.1 Gravitational force and field
State Newton’s universal law of gravitation.
The gravitational force is the weakest of the four fundamental forces, as the following visual shows:
ELECTRO-WEAK
STRONG
ELECTROMAGNETIC
WEAK
GRAVITY
© 2006 By Timothy K. Lund + +
light, heat, charge and magnets
nuclear
force
radioactivity
freefall, orbits
STRONGEST
WEAKEST

3. Topic 6: Fields and forces 9.2.1 Gravitational force and field
State Newton’s universal law of gravitation.
In the 1680’s in his groundbreaking book Principia Sir Isaac Newton published not only his works on physical motion, but what has been called by some the greatest scientific discovery of all time, his universal law of gravitation.
The law states that the gravitational force between two point masses m1 and m2 is proportional to their product, and inversely proportional to the square of their separation r.
The actual value of G, the gravitational constant, was not known until Henry Cavendish conducted a tricky experiment in 1798 to find it.
© 2006 By Timothy K. Lund
F = Gm1m2/r2
Universal law of gravitation
where G = 6.67×10−11 N m2 kg−2

4. Topic 6: Fields and forces 9.2.1 Gravitational force and field
State Newton’s universal law of gravitation.
Newton spent much time developing integral calculus to prove that a spherically symmetric shell of mass M acts as if all of its mass is located at its centre.
Thus the law works not only for point masses, which have no radii, but for any spherical distribution of mass at any radius like planets and stars.
- Newton’s shell theorem.
© 2006 By Timothy K. Lund M m r

5. Topic 6: Fields and forces 9.2.1 Gravitational force and field
You do not have to recall this!
Topic 6: Fields and forces Gravitational force and field
State Newton’s universal law of gravitation.
The earth has many layers, kind of like an onion:
Since each shell is symmetric, the gravi tational force caused by that shell acts as though it is all concentrated at its centre.
Thus the net force at m caused by the shells is given by
F = GMim/r2 + GMom/r2 + GMmm/r2 + GMcm/r2
F = G(Mi + Mo + Mm + Mc)m/r2
F = GMm/r2 where M = Mi + Mo + Mm + Mc
which is the total mass of the earth.
crust Mc
mantle Mm m
outer core Mo
inner core Mi r
© 2006 By Timothy K. Lund

6. Topic 9.2: Space 9.2.1 Gravitational force and field
State Newton’s universal law of gravitation.
Be very clear that r is the distance between the centres of the masses. m1 m2
F12
F21 r
EXAMPLE: The earth has a mass of M = 5.981024 kg and the moon has a mass of m = 7.361022 kg. The mean distance between the earth and the moon is 3.82108 m. What is the gravitational force between them?
SOLUTION:
F = GMm/r2
F = (6.67×10−11)(5.981024 )(7.361022 )/(3.82108)2
F = 2.011020 n.
© 2006 By Timothy K. Lund
* The radii of each planet is immaterial to this problem.

7. Topic 9.2: Space 9.2.1 Gravitational force and field
Define gravitational field strength.
Suppose a mass m is located a distance r from a another mass M.
We define the gravitational field strength g as the force per unit mass acting on m due to the presence of M. Thus
The units are newtons per kilogram (N kg-1).
Note that from Newton’s second law, F = ma, we see that a N kg-1 is also a m s-2, the units for acceleration.
Note further that weight has the formula F = mg, and the g in this formula is none other than the gravitational field strength!
On the earth’s surface, g = 9.8 N kg-1 = 9.8 m s-2.
g = F/m
gravitational field strength
© 2006 By Timothy K. Lund

8. Topic 9.2: Space 9.2.1 Gravitational force and field
Derive an expression for gravitational field strength at the surface of a planet assuming that all its mass is concentrated at its centre.
Suppose a mass m is located on the surface of a planet of radius R. We know that it’s weight is F = mg.
But from the law of universal gravitation, the weight of m is equal to its attraction to the planet’s mass M and equals F = GMm/R2.
Thus mg = GMm/R2.
This same derivation works for any r.
© 2006 By Timothy K. Lund
g = GM/R2
gravitational field strength at the surface of a planet of mass M and radius R
g = GM/r2
gravitational field strength at a distance ‘r’ from the centre of a planet

9. Topic 9.2: Space 9.2.1 Gravitational force and field
Determine the gravitational field due to one or more point masses.
PRACTICE: Given that the mass of the earth is M = 5.981024 kg and the radius of the earth is R = 6.37106 m, find the gravitational field strength at the surface of the earth, and at a distance of one earth radii above the surface.
SOLUTION:
For r = R:
g = GM/R2
g = (6.67×10−11)(5.981024)/(6.37106)2
g = 9.83 N kg-1 (m s-2).
For r = 2R: Since r is squared…
just divide by 22 = 4. Thus
g = 9.83/4 = 2.46 m s-2.
© 2006 By Timothy K. Lund
FYI
 A (N.kg-1) is the same as a (m.s-2)

10. Topic 9.2: Space 9.2.1 Gravitational force and field
Define gravitational field strength.
If we take a top view, and eliminate some of the field arrows, our sketch of the gravitational field is vastly simplified:
In fact, we don’t even have to draw the sun-the arrows are sufficient to denote its presence.
To simplify field drawings even more, we take the convention of drawing “field lines” with arrows in their centres.
SUN
© 2006 By Timothy K. Lund
SUN

11. Topic 9.2: Space 9.2.1 Gravitational force and field
Define gravitational field strength.
In the first sketch the strength of the field is determined by the length of the field arrow.
Since the second sketch has lines, rather than arrows, how do we know how strong the field is at a particular place in the vicinity of a mass?
We simply look at the concentration of the field lines. The closer together the field lines, the stronger the field.
In the red region the field lines are closer together than in the green region.
Therefore the red field is stronger than the green field.
SUN
SUN
© 2006 By Timothy K. Lund

12. Topic 6: Fields and forces 9.2.1 Gravitational force and field
Determine the gravitational field due to one or more point masses.
PRACTICE: Sketch the gravitational field about the earth (a) as viewed from far away, and (b) as viewed “locally” (at the surface).
SOLUTION:
(a)
(b) or
© 2006 By Timothy K. Lund
FYI
Note that the closer to the surface we are, the more uniform the field concentration.

13. Topic 9.2: Space 9.2.1 Gravitational force and field
Determine the gravitational field due to one or more point masses.
EXAMPLE: Find the gravitational field strength at a point between the earth and the moon that is right between their centers.
SOLUTION:
A sketch may help.
Let r = d/2. Thus
gm = Gm/(d/2)2
gm = (6.67×10−11)(7.361022)/(3.82108/2)2
gm = 1.3510-4 N.
gM = GM/(d/2)2
gM = (6.67×10−11)(5.981024)/(3.82108/2)2
gM = 1.0910-2 N. Thus g = gM – gm = 1.0810-2 N.
M = 5.981024 kg
m = 7.361022 kg gm gM
d = 3.82108 m
© 2006 By Timothy K. Lund

14. Topic 9.2: Space 9.2.1 Gravitational force and field
Determine the gravitational field due to one or more point masses.
EXAMPLE: Two masses of 225-kg each are located at opposite corners of a square having a side length of 645 m. Find the gravitational field vector at (a) the centre of the square, and (b) one of the unoccupied corners.
SOLUTION: Start by making a sketch.
(a) The opposing fields cancel so g = 0.
(b) The two fields are at right angles.
g1 = (6.67×10−11)(225)/(645)2 = 3.6110-14 N
g2 = (6.67×10−11)(225)/(645)2 = 3.6110-14 N
g2 = g12 + g22 = 2(3.6110-14)2 = 2.6110-27
g = 5.1110-14 N. m g2
(b) s g1
(a) m
© 2006 By Timothy K. Lund
g1+g2 g1 g2
sum points to centre of square

15. Topic 9.2: Space 9.2.1 Gravitational force and field
Determine the gravitational field due to one or more point masses.
PRACTICE: Determine the gravitational field strength at the points A and B.
SOLUTION: Organize masses and sketch fields.
For point A:
g1 = (6.67×10−11)(125)/(225)2 = 1.6510-13 N
g2 = (6.67×10−11)(975)/( )2 = 4.0610-13 N
g = g2 – g1 = 4.06 10-13 = 2.4110-13 N.
For point B:
g1 = (6.67×10−11)(125)/( )2 = 1.1510-14 N
g2 = (6.67×10−11)(975)/(225)2 = 1.2810-12 N
g = g1 + g2 = 1.15 10-12 = 1.2910-12 N.
975 kg
125 kg A B g1 g2 g2 g1 1
625 m 2
225 m
225 m
© 2006 By Timothy K. Lund

16. Topic 9.2: Space 9.2.1 Gravitational force and field
Solve problems involving gravitational forces and fields.
PRACTICE:
Jupiter’s gravitational field strength at its surface is 25 N.kg-1 while its radius is 7.1107 m.
(a) Derive an expression for the gravitational field strength at the surface of a planet in terms of its mass M and radius R and the gravitational constant G.
SOLUTION: This is for a general planet…
F = Gm1m2/r2 (law of universal gravitation)
F = GMm2/R2 (substitution)
g = F/m2 (definition of gravitational field)
g = (GMm2/R2)/m2 (substitution)
g = GM/R2
© 2006 By Timothy K. Lund

17. Topic 9.2: Space 9.2.1 Gravitational force and field
Solve problems involving gravitational forces and fields.
PRACTICE:
Jupiter’s gravitational field strength at its surface is 25 N kg-1 while its radius is 7.1107 m.
(b) Using the given information and the formula you just derived deduce Jupiter’s mass.
(c) Find the weight of a 65-kg man on Jupiter.
SOLUTION:
(b) g = GM/R2 (just derived in (a))
M = gR2/G (manipulation)
M = (25)(7.1107)2/6.67×10−11
M = 1.9×1027 kg.
(c) F = mg
F = 65(25) = 1600 N.
© 2006 By Timothy K. Lund

18. Topic 9.2: Space 9.2.1 Gravitational force and field
For Advanced Learner’s Only
Solve problems involving gravitational forces and fields.
PRACTICE: Two isolated spheres of equal mass and different radii are held a distance d apart. The gravitational field strength is measured on the line joining the two masses at position x which varies. Which graph shows the variation of g with x correctly?
There is a point between M and m where g = 0.
Since g = Gm/R2 and Rleft < Rright, gleft > gright at the surfaces of the masses.
© 2006 By Timothy K. Lund