1. Newton’s Law of Universal Gravitationby
Daniel Silver
AP Physics C
Chapter 11
Physics for Scientists and Engineers, 4th ed
Modified by Stephanie Ingle, January 2009

2. Universal Law of Gravitation
Every object attracts every other object with a force that is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between them.

3. Calculating Gravitational Force
Fg = gravitational force
w = weight
G = universal constant of gravitation (6.67E-11)
m1, m2 = masses of objects
d = distance from center to center between objects Fg Fg m1 m2 d

4. Orbital Velocity for a Satellite moving in a Circular Orbits
Centripetal Force = Gravitational Force v

5. Useful Constants G = 6.67E-11 Nm2/kg2 Mass of Earth = 5.98E24 kg
Radius of Earth = 6.4E6 m

6. Kepler’s Laws overview
1st Law: “Planets move in elliptical orbits with the sun at one focus of the ellipse.”
2nd Law: “A line from the sun to a given planet sweeps out equal areas in equal times.”
3rd Law: The cube of the ratio of the orbital radius is equal to the square of the ratio of the period of rotation

7. Kepler’s 1st Law – Elliptical Orbits
“Planets move in elliptical orbits with the sun at one focus of the ellipse.”
Eccentricity is a property of an ellipse that describes how elongated the ellipse is.
Sun
Empty focus
Planet
Minor axis
Major axis b a
a = semi-major axis
b = semi-minor axis

8. Satellite Motion w/ Keplers 1st Law
Planetary motion is just one example of satellite motion…therefore this law can be extended to any satellite orbiting around any body.
Circular orbits are a special case of an elliptical orbit. (eccentricity =0)
Elliptical orbits have an eccentricity between zero and one…the larger the number, the more elongated the ellipse.

9. Kepler’s 2nd Law – Equal Areas in Equal Time
“A line from the sun to a given planet sweeps out equal areas in equal times.” S a b d c
This law helps to illustrate that planets (or satellites) in elliptical orbits travel faster when they are closer to the Sun (or body they are orbiting.
If a planet takes 20 days to move from point “a” to point “b” and also moves from point “c” to point “d” in the same amount of time…the areas of aSb and cSd will be equal.

10. Kepler’s 3rd Law – Orbital period and distance related
The cube of the ratio of the orbital radius is equal to the square of the ratio if the period of rotation
So for a satellite moving in a circular orbit… v
and
With a little substitution and algebra, we can solve for the orbital period. or

11. Kepler’s 3rd Law – Elliptical Orbits
For elliptical orbits there is not a constant orbital radius so we must use the semi-major axis “a” in place of the “r” for orbital distance. v or

12. Gravitational Potential Energy The Physics 1 way – “local Ug”
So far, we have defined gravitational potential energy (Ug) relative a reference point at the lowest point the object can go (often the ground) which is given a value of zero.
In doing this we have been able to calculate gravitational potential energy simply in terms of height above the lowest possible point (or surface of the earth), which is great for conserving energy “locally” or near the surface of the earth.
In fact, the formula:
Assumes the use of the local acceleration due to gravity “g” at the surface of the earth, indicating that it is only valid for energy conservation near the surface of the earth.

13. Gravitational Potential Energy
With Calculus and anywhere in the universe
Now that we are defining gravitation in a more universal sense, we must re-define the zero reference point to be infinitely away from the earth’s center. In otherwords… at
Recall that force and potential energy are related …
If we rearrange this equation and do some substitution and algebra…
Where “r” is a position vector measured radially rather than along an x-, y-, z- axis.

14. Gravitational Potential Energy at the Earth’s Surface
Earth, ME RE m
An object of mass “m” has a Gravitational Potential Energy of

15. Conservation of Energy in a Gravitational Field
Earth, ME m h
If an object (i.e. rocket) of mass “m” were launched from the earth’s surface at velocity “vi”, calculate the velocity “vf” at a height “h” above the earth’s surface. RE
Start by conserving energy the same way as before… K1+U1=K2+U2 … then substitute the equations for K & U at each point…then solve for “vf”

16. Escape Velocity m
Escape velocity is defined as the minimum velocity an object must have in order to completely escape the planet’s (earth’s) gravitational pull.
According to the potential energy formula…
The planet’s gravitational potential energy approaches zero as “r” approaches ∞, so we need to give our rocket exactly the amount of Kinetic energy to start with so that it will use it all up to get to ∞. In other words, at r= ∞, Ug=0 and K=0. To calculate the escape velocity from the earth’s surface… m RE
So escape velocity from the earth’s surface is …
Earth, ME