Download The Newton`s law of gravitation.

PHYS1110H, 2011 Fall.
Shijie Zhong
Gravity
The Newton’s law of gravitation.
For two objects with masses M and m at a distance of r, the gravitational
attraction force is in terms of magnitude
GMm
,
(1)
F=
r2
where G is the universal gravitational constant and 6.67x10-11 N m2/kg2.
The force is attractive and the direction of the force to mass M points to
mass m and vice verse.
€
If a coordinate system is set up such that its origin is at the center of mass M,
the force on mass m is given

GMm
(2)
F =−
eˆr ,
r2
where eˆr is an unit vector pointing from M to m, which is the reason for the
negative sign.
€
€
Remarks:
1) Gravitational forces are very weak.
2) The force acting on m from M is the same as the force acting on M
from m in terms of the magnitude but opposite in directions -Newton’s 3rd law.
Gravitational potential energy.
€
As discussed in note 5, gravitational force is a conservative force for which
gravitational potential energy can be defined. The work done by the
gravitational force to move mass m from p1 to p2 is (independent of path).
p2   GMm GMm
W1−2 = ∫ F ⋅ dr =
−
= −(U 2 −U1 ) ,
(3)
r2
r1
p1
GMm
U(r) = −
.
(4)
r
This potential energy represents a potential energy “well” that bounds mass
m around mass M.
€
Note
6
1
€
€
€
Example 1. Energy of a satellite
Find out the energy of a satellite with m that orbits the Earth at a distance of
r from the center of the Earth. Earth’s mass is M.
1
GMm
E = KE + U = mv2 −
.
(5)
2
r
The velocity of the satellite v needs to satisfy the Newton’s law:
GM
GMm
v2
v2 =
or
.
(6)
=m ,
r
r
r2
GMm
E =−
.
(7)
2r
The magnitude of E is always half of the gravitational potential energy, and
€
E is less negative than U hence larger than U (due to KE). E<0, so the
satellite is still bounded by the Earth’s gravitational energy, i.e., remains as
Earth’s satellite!
A synchronous satellite is a satellite that orbits the Earth at the same angular
velocity as the Earth, i.e., always hanging above the same location on the
Earth. What’s its orbital radius?
(6) is always true, and now an additional constraint is v = rω , where ω is the
rotation rate of the earth, or 2π/day ~ 2x10-7 /second. Combining these two
constraints, we have
GM
€
.
(8)
r=3
2
ω
r does not depend on the mass of satellite.
€
€
Example 2. Escape velocity
Find out the escape velocity of an object on the surface of Earth. Earth’s
mass is M, and radius is RE. Note that we solved this problem in note 4’s
example 5 using the work-energy theorem.
For an object at the surface of Earth with speed vs, its total energy at the
surface of the Earth is:
1
GMm
,
(9)
E s = KE s + U s = mvs2 −
2
RE
The conservation of energy says that when the object is at some radius r, its
total energy remains as Es. For the object to escape the Earth’s gravitational
field, the mass m must have a total energy E=0 to get out the potential well.
1
GMm
(10)
E s = mvs2 −
= 0,
2
RE
Note
6
€
2
This leads to vs = 2GM / R E , which is velocity needed to escape from the
gravitational field of the Earth. This is the same answer as we got before.
€
Gravitational acceleration g
The gravitational attraction force on mass m at a distance r from the center
of the Earth is given by (2). Suppose that the gravitational force is the only
force on mass m (i.e., in the air) and that the mass is in a free fall. According
to Newton’s 2nd law:

GMm

(11)
F =−
eˆr = ma ,
r2
GM


or a = −
(12)
eˆr = g ,
r2
This acceleration is called gravitational acceleration g.
€
€
Remarks:
1) On the surface of the Earth, g=GM/RE2. Using the mass and radius of
the Earth, g is found to be 9.8 m/s2. It is only determined by the mass
and radius of the Earth. This is why a metal ball or a piece of paper all
experiences the same gravitational acceleration.
2) On the surface of the Moon, g is 1.63 m/s2 or ~1/6 of that on the
Earth’s surface, because the Moon is less massive.
3) At a typical altitude of space shuttle (h~300 km above the ground),
g= GM/(RE+h)2, which is only ~10% smaller than g on the surface of
the Earth. However, we often hear about astronaut being “weightless”
in the space shuttle. Why? The answer is the circular motion of the
space shuttle around the Earth that prevents the space shuttle and
astronauts from falling to the ground.
Note
6
3
Kepler’s laws on orbital motion.
By summarizing astronomical observations made by his mentor Brahe,
Kepler came up the following three laws:
1) Each planet moves in an ellipse with the Sun at one focus.
2) The radius vector from the Sun to a planet sweeps out equal areas in
equal times.
3) The period of revolution T of a planet around the Sun is related to the
major axis of the ellipse A by
2
T = kA 3,
(13)
where k is a constant.
€
€
For now, we limit ourselves to circular orbits which are special cases of
elliptic orbits. For circular orbits, Kepler’s 1st and 2nd laws are obvious. The
third law can be derived from Newton’s gravitational law – actually Kepler’s
laws really motivated Newton’s law of gravitation and Newtonian
mechanics!
From Newton’s 2nd law and Newton’s law of gravitation, for the circular
motion of a planet around the Sun with orbital radius r, we have
GM s m p
GM s
v2
v2 =
or
.
(14)
= mp ,
r
r
r2
For circular motion with a period of T, v=2πr/T, and substituting it to (14)
leads to
4π 2 3
2
€(15)
T =
r .
GM s
€
Note
6
4