GRAVITY

1. Newton's law of universal gravitation states that the force of attraction between two
masses and is and is directed along the line joining the two bodies.
Putting in a constan
m m F mm
r
∝
1 2
2
21 2 1
12 1 2 12 21
t of proportionality, . Now let's be a bit careful of the
direction of the force. Looking at the diagram below, Force on by ,
Force on by , . By New
F Gmm
r
F m m
F m m F F F
=
=
= = =
􀁇
􀁇 􀁇 􀁇
12 21 ton's Third Law, F = −F.
􀁇 􀁇
2. The gravitational constant G is a very small quantity and
needs a very sensitive experiment. An early experiment to
find involved suspending two masses and measuring the
attrac
G
2
tive force. From the figure you can see that the
is gravitational torque is 2 . A thread provides
2
the restoring torque . The deflection can be measured
by observing th
GmM L
r
κθ θ
⎛ ⎞
⎜ ⎟
⎝ ⎠
2
2
e beam of the light reflected from the small
mirror. In equilibrium the torques balance, .
Hence . How to find ? It can be found from
observing the period of free o
GmML
r
G r
GmML
κθ
κθ
κ
=
=
2
2
2
11 2 2
scillations, 2 4
with . The modern value is 6.67259 10 . /
2
T I I
T
I mL G Nm kg
π
π κ
κ
−
= ⇒ =
= = ×
E
2 E E
E
3. The magnitude of the force with which the Earth attracts a body of mass m towards its
centre is , where 6400 is the radius of the Earth and is the mass.
The material does n
F GmM R km M
R
= =
E
2
E
ot matter - iron, wood, leather, etc. all feel the force in proportion to
their masses. If the body can fall freely, then it will accelerate. So, F mg GmM .
R
= =
1 m 2 m
r
12 F
􀁇
21 F
􀁇
mirror
scale
Quartz fiber m
M
θ
m
M
We measure g, the acceleration due to gravity, as 9.8m/s . From this we can immediately
deduce the Earth's mass: 5.97 10 . What a remarkable achievement!
We can do still more: th
E
E
M gR kg
G
= = ×
3 21 3
E E
E 3
E
E
e volume of the Earth 4 1.08 10 . Hence the]
3
density of the Earth 5462kg . So this is 5.462 times greater than the
density of water and tells us that the earth must be quit
V R m
V m
M
π
ρ −
= = = ×
= = =
e dense inside.
4. The is an important quantity. It is the work done in moving a unit
mass from infinity to a given point R, and equals ( ) .
Proof: Conservation of
V r GM
R
= −
gravitational potential
0
( )
2
energy says, ( )
Integrate both sides: 0 ( ) 1 , ( )
5. Using the above formula, let us calculate the change in potential energy when we
V R R
R R
dV Fdr dV drF r
V R GM dr GM V R GM
r r R
U
∞
∞ ∞
= − ⇒ = −
− = =− ⎡⎢⎣ ⎤⎥⎦ ∴ =−
Δ
∫ ∫
∫
( ( )1)
raise
a body of mass to a height above the Earth's surface.
1 1 1 1 1 1 /
1 /
Now suppose that the distance is much smalle
E
E E E
m h
U GMm GMm GMm h R
R R h h R
h
⎛ ⎞ ⎛ ⎞ −
Δ = ⎝⎜ − + ⎠⎟= ⎜⎝ − + ⎟⎠ = − +
( )1 ( ( ))
r than the Earth's radius. So, for ,
1 / 1 / . So we find 1 1 / .
6. We can use the expression for potential energy and the law of conservation of energy to
E
E E E
E
h R
h R h R U GMm h R m GM h mgh
R
− ⎛ ⎞
+ = − Δ = − − = ⎜ ⎟ =
⎝ ⎠
􀀓
find the minimum velocity needed for a body to escape the Earth' gravity. Far away from
the Earth, the potential energy is zero, and the smallest value for the kinetic energy is
zero. Requiring ( ) ( ) 2
e e
e
that gives 1 v 0 0. From
2
this, v 2 2 . Putting in some numbers we find that for the Earth v 11.2km/s
and for the Sun v 618km/s. For a Black Hole, the escap
r R r e
E
E
E
KE PE KE PE m GMm
R
GM gR
R
= =∞ + = + − = +
= = =
= e velocity is so high that nothing
can escape, even if it could move with the speed of light! (Nevertheless, Black Holes can be
observed because when matter falls into them, a certain kind of radiation is emitted.)
PHYSICS –PHY101 VU
© Copyright Virtual University of Pakistan
60
M θ r
Δθ rΔθ
ΔA
2
2 o
7. : A satellite is in circular orbit over the Earth's surface. The condition
for equilibrium, v v . If is the Earth's radius, and is the
height of the satel
o
E
m GMm GM R h
r r r
= ⇒ =
Satellite problems
o
o
3 / 2
o
lite above the ground, then . Hence, v . For
, we can approximate v . We can easily calculate the time
for one complete revolution, 2 2 2 2 . This g
v
E
E
E E
E
r R h GM
R h
h R GM gR
R
T r r r r
GM GM
π π π
π
ω
= + =
+
<< = =
= = = =
2
2 3 2 3
ives the
important result, observed by Kepler nearly 3 centuries ago that 4 , or .
8. What is the total energy of a satellite moving in a circular orbit around the earth? Clearly,
i
T r T r
GM
π
= ∝
2 2
t has two parts, kinetic and potential. Remeber that the potential energy is negative. So,
1 v . But, v as we saw earlier and therefore,
2
1 1 . Note that t
2 2
E
o o
E KE PE m GM m GM
r r
E GMm GMm GMm
r r r
= + = − =
= − =− he magnitude of the potential energy is
larger than the kinetic energy. If it wasn't, the satellite would not be bound to the Earth!
9. A famous discovery of the astronomer Johann Kepler some 300 years ago says that the
line joining a planet to the Sun sweeps out equal areas in equal intervals of time. We can
easily see this from the conservation of angular momentum. Call ΔA the area swep
( )
0
t out
in time . Then from the diagram below you can see that 1 . Divide this by
2
and then take the limit where it becomes very small,
lim 1
t 2
t Ar r
t
d A A r
dt t
θ
Δ →
Δ Δ= Δ
Δ
Δ
= =
Δ
2 2
0
lim 1 .
2 2
Since is a constant, we have proved one of Kepler's laws (with so little effort)!
t
r L
t m
L
θ
ω
Δ →
⎛ Δ ⎞
⎜⎝ Δ ⎟⎠= =