Download Review of Universal Gravitation

AP Physics C
Mrs. Coyle
Review of Universal Gravitation
-Newton’s Law of Universal Gravitation :
F=
𝑮𝒎𝟏 𝒎𝟐
𝒓𝟐
Where G= 6.67 x 10-11 N m2 / kg2,
r is the distance between the objects from center to center.
For multiple masses acting on one mass, use vector addition to find the resultant force vector
(principal of superposition).
For planets, assuming a near circular orbit, the gravitational force plays the role of the centripetal
force, so:
F=
𝑮𝒎𝒑 𝒎𝒔
𝒓𝟐
= mp
𝒗𝟐
𝒓
, where mp is the mass of the planet and ms is the mass of the sun.
-Gravitational Force as a function of distance from the center of the earth.
For r>R Earth the gravitational force acting on an object m by the earth is F=
𝑮𝑴𝒎
.
𝒓𝟐
For r<REarth only the portion of the earth’s mass that is enclosed by the imaginary surface causes a
force.
4
MEarth enclosed = ρ Venclosed = ρ 𝑉 = 𝜌 3 𝜋𝑟 3
F=
𝟒
𝟑
𝒓𝟐
𝑮 𝒎 𝝆 𝝅𝒓𝟑
=
𝟒 𝝅𝝆𝑮 𝒓
𝟑
, therefore the force acting
on an object inside the surface of the
earth is directly proportional to r.
Note: An object in an
imaginary underground
tunnel along the earth’s
diameter undergoes simple
harmonic motion. That’s
because F=-kr
𝑘
𝑘
𝑚
𝑚
a=- r, and then set 𝜔2 =
2𝜋
to find T=
𝜔
-Gravitational Potential Energy of a System of Masses
a) Gravitational Potential Energy for a Two Particle System:
U= -
𝑮𝑴𝒎
𝒓
Note: U= 0 when r = ∞
b) Gravitational Potential Energy for a System of More Than Two Particles:
U= -[
𝑮𝒎𝟏 𝒎𝟐
𝒓𝟏𝟐
+
𝑮𝒎𝟏𝒎𝟑
𝒓𝟏𝟑
+
𝑮𝒎𝟐𝒎𝟑
𝒓𝟐𝟑
]
Note: An external force need to do positive work in order to separate the particles.
-Potential Energy and Force
F= -
𝒅𝑼
𝒅𝒓
=-
𝒅 (
−𝑮𝑴𝒎
)
𝒓
𝒅𝒙
=-
𝑮𝒎𝑴
𝒓𝟐
Also:
W=∫ 𝑭 ∙ 𝒅𝒓
W= - ∆U
-Escape Speed
Minimum initial speed that when a satellite is shot from the surface of the earth it will come to rest
at infinity. Assume the radius of the earth is r and its mass M.
To find the escape velocity use conservation of energy:
Energy Initial =Energy Final
½ mv2 + ( -
𝑮𝑴𝒎
v= √
𝒓
)=0+0
𝟐𝑮𝑴
𝒓
-Kepler’s Laws
1st ) Law of Orbits (Ellipses).
All planets move in elliptical orbits with the sun at one focus.
2nd) Law of Areas.
A line that connects a planet and the Sun sweeps out equal areas in equal intervals of time.
This law is a statement of the law of conservation of angular momentum.
3rd) Law of Periods (Law of Harmonies).
𝑻𝟏 𝟐
𝒓𝟏 𝟑
[ ] =[ ]
𝑻𝟐
𝒓𝟐
This is proven by using F=m a= m(𝜔2 𝑟)
And equating it to F=
It results to T2 = (
4 𝜋2
𝐺𝑀
𝑮𝒎𝟏 𝑴
𝒓𝟐
) 𝑟 3, then take the ratios for two different satellites or planets.
- Energies of Satellite-Planet System
a) Circular Orbit
Energy =
E= ½ mv2 + ( -
𝑮𝑴𝒎
𝒓
)
Since the orbit is circular F=
E=
𝑮𝒎𝑴
𝟐𝒓
+(-
𝑮𝑴𝒎
𝒓
𝑮𝒎𝑴
𝒓𝟐
=
𝒎𝒗𝟐
𝒓
 ½ mv2
=
𝑮𝒎𝑴
𝟐𝒓
)
E=-
𝑮𝒎𝑴
𝟐𝒓
= -K
So, for a circular orbit E = - K
b) Elliptical Orbit
Replace the r with the semimajor axis α.
E=-
𝑮𝒎𝑴
𝟐𝒂
(Note that the energy does not depend on the eccentricity)
Note: For a system of two bound particles both the total energy and the total angular
momentum are conserved.
Examples:
1) How does the speed of a satellite at position 2 compare to the speed at position 1. The
distance r2 =2r1. (Hint: Use conservation of angular momentum)
2
1
E
a
r
t
h
2) When a satellite is in circular orbit and momentarily burns the rocket perpendicular to the radius
vector in such a way that the speed of the rocket increases. What is the new orbit?
Answer:
The kinetic energy of the satellite will increase. The potential energy will remain the same because
the burn is quick and we can assume that the rocket did not change position during the burn. So the
total energy E new > E old (Enew will be less negative) . Therefore, the satellite will no longer follow
equation (1) and will not be in circular orbit at that same instantaneous radius.
Instead the satellite will follow an elliptical orbit where from equation (4), since E new is less
negative, the absolute value of E will be smaller, a will be larger than r. The path followed will be
an ellipse as shown (the orbit where vCircular< v < vEscape )
Other possible orbits are achieved by varying the velocity of the satellite as shown above.