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Gravity and Satellite Motion: Understanding Newton's Laws and Kepler's Laws, Lecture notes of Classical Physics

University of AllahabadClassical Physics

An in-depth analysis of newton's law of universal gravitation and its application to satellite motion. It covers the concepts of gravitational force, potential energy, and conservation of energy. Additionally, it explains kepler's laws of planetary motion derived from the conservation of angular momentum.

Typology: Lecture notes

2011/2012

Uploaded on 08/12/2012

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bg1
PHYSICS –PHY101 VU
© Copyright Virtual University of Pakistan
58
Summary of Lecture 21 – GRAVITY
12
12 2
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
mm
mmF
r
∝
12
2
21 2 1
12 1 2 12 21
t of proportionality, . Now let's be a
b
it careful of the
direction of the force. Looking at the diagram below, Force on by ,
Force on by , . By New
mm
FG
r
Fmm
FmmFFF
=
=
===

12 21
ton's Third Law, .FF=−

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
r
GGmML
κθ
κθ κ
=
=
2
2
2
11 2 2
4
scillations, 2
with . The modern value is 6.67259 10 . /
2
I
I
TT
mL
I
GNmkg
π
πκ
κ
−
=⇒=
==×
E
EE
2
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
GmM
FRkm 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, .
GmM
Fmg R
==
1
m2
m
r
12
F
21
F
mirror
scale
Quartz fiber m
M
θ
m
M
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Download Gravity and Satellite Motion: Understanding Newton's Laws and Kepler's Laws and more Lecture notes Classical Physics in PDF only on Docsity!

Summary of Lecture 21 – GRAVITY

1 2 (^1 2 )

  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 m m F 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

m m F G r

F m m

F m m F F F

G

G G G

ton's Third Law, F 12 (^) = − F 21.

G G

  1. 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

r G GmML

κθ

κθ κ

2

2

2 11 2 2

scillations, 2

with. The modern value is 6.67259 10. / 2

I I

T

T

mL I G N m kg

π π κ κ

−

= = ×

E 2 E E E

  1. 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

GmM F 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,.

GmM F mg R

m 1 m 2

r

F 12

G

F 21

G

mirror

scale

Quartz fiber

m

M

θ

m

M

2

2 24

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

gR M kg G

= = ×

3 21 3 E E

E^3 E E

e volume of the Earth 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.

  1. 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

GM

V r R

gravitational potential

0

( )

2

energy says, ( )

Integrate both sides: 0 ( ) , ( )

  1. Using the above formula, let us calculate the change in potential energy when we

V R R

R R

dV Fdr dV drF r

dr GM V R GM GM V R r r R

U

∞

∞ ∞

( (^ ) )

1

raise

a body of mass to a height above the Earth's surface.

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

⎝ +^ ⎠ ⎝ + ⎠

1

r than the Earth's radius. So, for ,

1 / 1 /. So we find 1 1 /.

  1. We can use the expression for potential energy and the law of conservation of energy to

E

E E E E

h R

GM

h R h R U GMm h R m h mgh R

− ⎛^ ⎞

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 v 0 0. From 2

this, v 2. Putting in some numbers we find that for the Earth v 11.2 km/s

and for the Sun v 618km/s. For a Black Hole, the escap

r R r^ e E

E E

GMm KE PE KE PE m 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.)