B.Sc. Sem-I Paper-I

1. Shri Govindrao Munghate Arts and Science College ,
Kurkheda
Department of Physics
Topic –Gravity
Prof. B.V. Tupte
Head, Department of Physics
Shri Govindrao Munghate Arts and Science
College Kurkheda.
Newton’s Law of Gravitation
Kepler’s Laws of Planetary Motion
Gravitational Fields

2. Newton’s Law of Gravitation
Kepler’s Laws of Planetary
Motion
Gravitational Fields

3. Newton’s Law of Gravitation
m1
m2
r
There is a force of gravity between any pair of objects
anywhere. The force is proportional to each mass and
inversely proportional to the square of the distance between
the two objects. Its equation is:
FG =
G m1 m2
r2
The constant of proportionality is G, the universal gravitation
constant. G = 6.67 · 10-11 N·m2 / kg2. Note how the units of G all
cancel out except for the Newtons, which is the unit needed on the
left side of the equation.

4. Gravity Example
FG =
G m1 m2
r2
How hard do two planets pull on each other if their masses are
1.231026 kg and 5.211022 kg and they 230 million kilometers
apart?
This is the force each planet exerts on the other. Note the denominator
is has a factor of 103 to convert to meters and a factor of 106 to
account for the million. It doesn’t matter which way or how fast the
planets are moving.
(6.67 · 10-11 N·m2 / kg2) (1.23 · 1026 kg) (5.21 · 1022 kg)
=
(230 · 103 · 106 m)2
= 8.08 · 1015 N

5. 3rd Law: Action-Reaction
In the last example the force on each planet is the same. This is due to
to Newton’s third law of motion: the force on Planet 1 due to Planet 2
is just as strong but in the opposite direction as the force on Planet 2
due to Planet 1. The effects of these forces are not the same, however,
since the planets have different masses.
For the big planet: a = (8.08 ·1015 N) / (1.23 ·1026 kg)
= 6.57 ·10-11 m/s2.
For the little planet: a = (8.08 ·1015 N) / (5.21 · 1022 kg)
= 1.55 ·10-7 m/s2.
5.21
· 1022 kg
1.23 ·1026 kg
8.08 ·1015 N 8.08 ·1015 N

6. Inverse Square Law
FG =
G m1 m2
r2
The law of gravitation is called an inverse square law because the
magnitude of the force is inversely proportional to the square of the
separation. If the masses are moved twice as far apart, the force of
gravity between is cut by a factor of four. Triple the separation and
the force is nine times weaker.
What if each mass and the separation were all quadrupled?
answer: no change in the force

7. Calculating the Gravitational Constant
In 1798 Sir Henry Cavendish suspended a rod with two small masses
(red) from a thin wire. Two larger mass (green) attract the small
masses and cause the wire to twist slightly, since each force of
attraction produces a torque in the same direction. By varying the
masses and measuring the separations and the amount of twist,
Cavendish was the first to calculate G.
Since G is only
6.67 · 10-11 N·m2 / kg2,
the measurements had
to be very precise.

8. Calculating the mass of the Earth
Knowing G, we can now actually calculate the mass of the Earth. All
we do is write the weight of any object in two different ways and
equate them. Its weight is the force of gravity between it and the
Earth, which is FG in the equation below. ME is the mass of the
Earth, RE is the radius of the Earth, and m is the mass of the object.
The object’s weight can also be written as mg.
=
FG =
G m1 m2
r2
G ME m
RE
2 = mg
The m’s cancel in the last equation. g can be measured experimentally;
Cavendish determined G’s value; and RE can be calculated at
6.37 · 106 m (see next slide). ME is the only unknown. Solving for
ME we have:
g RE
2
G
ME = = 5.98 ·1024 kg

9. Calculating the radius of the Earth


RE
This is similar to the way the Greeks approximated Earth’s radius over
2000 years ago:
s
 is also the central angle of the arc. s = RE
RE = s /  6.37 ·106 m
Earth

10. Net Force Gravity Problem
3106 kg
2.7106 kg 2.5106 kg
40 m
60 m
3 asteroids are positioned as shown,
forming a right triangle. Find the net
force on the 2.5 million kg asteroid.
Steps:
1. Find each force of gravity on it and draw in the vectors.
2. Find the angle at the lower right.
3. One force vector is to the left; break the other one down into
components.
4. Find the resultant vector: magnitude via Pythagorean theorem;
direction via inverse tangent.
answer: 0.212 N at 14.6 above horizontal (N of W)

11. Falling Around
the Earth v
Newton imagined a cannon ball fired
horizontally from a mountain top at a speed
v. In a time t it falls a distance y = 0.5 g t2 while
moving horizontally a distance x = vt. If fired fast
enough (about 8 km/s), the Earth would curve downward
the same amount the cannon ball falls downward. Thus, the
projectile would never hit the ground, and it would be in orbit.
The moon “falls” around Earth in the exact same way but at a
much greater altitude. .
x = vt
y = 0.5 g t 2 {
continued on next slide

12. Necessary Launch Speed for Orbit
R = Earth’s radius
t = small amount of time after launch
x = horiz. distance traveled in time t
y = vertical distance fallen in time t
R
R
y = gt 2 / 2
x = vt
(If t is very small, the red
segment is nearly vertical.)
x2 + R2 = (R + y)2
= R2 + 2Ry + y2
Since y << R,
x2 + R2  R2 + 2Ry
 x2  2Ry
 v2 t2  2R(g t2 / 2)
 v2  R g. So,
v  (6.37 ·106 m ·9.8 m/s2)
½
v  7900 m/s

13. Early Astronomers
In the 2nd century AD the Alexandrian astronomer Ptolemy put forth a
theory that Earth is stationary and at the center of the universe and that
the sun, moon, and planets revolve around it. Though incorrect, it was
accepted for centuries.
In the early 1500’s the Polish astronomer Nicolaus
Copernicus boldly rejected Ptolemy’s geocentric model
for a heliocentric one. His theory put the sun stated that
the planets revolve around the sun in circular orbits and
that Earth rotates daily on its axis.
In the late 1500’s the Danish astronomer Tycho Brahe
made better measurements of the planets and stars than
anyone before him. The telescope had yet to be
invented. He believed in a Ptolemaic-Coperican
hybrid model in which the planets revolve around the
sun, which in turn revolves around the Earth.

14. Early Astronomers
In the late 1500’s and early 1600’s the Italian scientist
Galileo was one of the very few people to advocate the
Copernican view, for which the Church eventually had
him placed under house arrest. After hearing about the
invention of a spyglass in Holland, Galileo made a
telescope and discovered four moons of Jupiter, craters
on the moon, and the phases of Venus.
The German astronomer Johannes Kepler was a
contemporary of Galileo and an assistant to Tycho
Brahe. Like Galileo, Kepler believed in the heliocentric
system of Copernicus, but using Brahe’s planetary data
he deduced that the planets move in ellipses rather than
circles. This is the first of three planetary laws that
Kepler formulated based on Brahe’s data.
Both Galileo and Kepler contributed greatly to work of the English
scientist Sir Isaac Newton a generation later.

15. Kepler’s Laws of
Planetary Motion
1. Planets move around the sun in elliptical paths with the
sun at one focus of the ellipse.
2. While orbiting, a planet sweep out equal areas in equal
times.
3. The square of a planet’s period (revolution time) is
proportional to the cube of its mean distance from the sun:
T2  R3
Here is a summary of Kepler’s 3 Laws:
These laws apply to any satellite orbiting a much
larger body.

16. Kepler’s First Law
Planets move around the sun in elliptical paths with the
sun at one focus of the ellipse.
An ellipse has two foci, F1 and F2. For any point P on the ellipse,
F1 P + F2 P is a constant. The orbits of the planets are nearly circular
(F1 and F2 are close together), but not perfect circles. A circle is a an
ellipse with both foci at the same point--the center. Comets have very
eccentric (highly elliptical) orbits.
F1 F2
Sun
Planet
P

17. Kepler’s Second Law
Sun
While orbiting, a planet sweep out equal areas in equal times.
C
D
A
B
The blue shaded sector has the same area as the red shaded sector.
Thus, a planet moves from C to D in the same amount of time as it
moves from A to B. This means a planet must move faster when it’s
closer to the sun. For planets this affect is small, but for comets it’s
quite noticeable, since a comet’s orbit is has much greater eccentricity.
(proven in advanced physics)

18. Kepler’s Third Law
The square of a planet’s period is proportional to the
cube of its mean distance from the sun: T 2  R3
Assuming that a planet’s orbit is circular (which is not exactly correct
but is a good approximation in most cases), then the mean distance
from the sun is a constant--the radius. F is the force of gravity on the
planet. F is also the centripetal force. If the orbit is circular, the
planet’s speed is constant, and v = 2R / T. Therefore,
Sun
Planet
F
R
M
m
G M m
R2
m v2
R
=
G M
R2
m [2R / T]2
R
=
Cancel m’s
and simplify:
42R
T 2
=
Rearrange:
G M
T 2
42
= R3
Since G, M, and  are constants, T 2  R3.

19. Third Law Analysis
GM
T 2
42
= R3
We just derived
• It also shows that the orbital period depends on the mass of the
central body (which for a planet is its star) but not on the mass of the
orbiting body. In other words, if Mars had a companion planet the same
distance from the sun, it would have the same period as Mars,
regardless of its size.
• This shows that the farther away a planet is from its star, the longer it
takes to complete an orbit. Likewise, an artificial satellite circling Earth
from a great distance has a greater period than a satellite orbiting closer.
There are two reasons for this: 1. The farther away the satellite is, the
farther it must travel to complete an orbit; 2. The farther out its orbit is,
the slower it moves, as shown:
G M m
R2
m v2
R
=  v =
G M
R

20. Third Law Example
One astronomical unit (AU) is the distance between Earth and the
sun (about 93 million miles). Jupiter is 5.2 AU from the sun. How
long is a Jovian year?
answer: Kepler’s 3rd Law says T 2  R3, so T 2 = k R3, where k
is the constant of proportionality. Thus, for Earth and Jupiter we
have:
TE
2 = k RE
3 and TJ
2 = k RJ
3
k’s value matters not; since both planets are orbiting the same central
body (the sun), k is the same in both equations. TE = 1 year, and
RJ /RE = 5.2, so dividing equations:
TJ
2
TE
2
RJ
3
RE
3
=  TJ
2 = (5.2)3
 TJ = 11.9 years
continued on next slide

21. 1 year
365 days
Third Law Example (cont.)
What is Jupiter’s orbital speed?
answer: Since it’s orbital is approximately circular, and it’s speed is
approximately constant:
2 (5.2)(93·106 miles)
v =
d
t
=
11.9 years
Jupiter is 5.2 AU from the sun (5.2
times farther than Earth is).
· ·
1 day
24 hours
 29,000 mph. Jupiter’s period from last slide
This means Jupiter is cruising through the solar system at about
13,000 m/s! Even at this great speed, though, Jupiter is so far away
that when we observe it from Earth, we don’t notice it’s motion.
Planets closer to the sun orbit even faster. Mercury, the closest
planet, is traveling at about 48,000 m/s!