1. I.B. Gravity of a Solid Sphere
Graph of |Fg| vs. r for a solid sphere of uniform density r:
|Fg(inside)| = Gm(r)m/ r2 = G{(4p/3)rr3}/r2 ~ r.
|Fg(outside)| = GMEm/ r2 = ~ r-2.
|Fg| R
~ 1/r2
~ r r
(I.B.1)
(I.B.2)

2. I.B. Gravity of a Solid Sphere
2. Graph of |Fg| vs. r for a solid sphere with M(r) = Mtotr2
|Fg(inside)| = Gm(r)m/ r2 = G{Mtotr2}/r2 ~ constant.
|Fg(outside)| = GMEm/ r2 = ~ r-2.
r ~ r-1.
|Fg| R
~ 1/r2
~ r r
(I.B.3)
(I.B.4)

3. I.B. Gravity of a Solid Sphere
Inside a sphere, form of gravitational force depends on the distribution of matter--the density.
M(r) = ∫rdV. (I.B.5)
4. Outside a sphere, form of gravitational force depends on the inverse square law.
5. Far enough away from any finte matter distribution, |Fg|~ r-2.
At Earth’s surface: Fg = GmEm/RE2
This is just the weight of mass m at the surface: W = Fg = GmEm/RE2 = mg, where
g = GmE/RE2. (I.B.6)

4. I.C. Gravitational Potential Energy
For uniform gravity, U = mgy (e.g., near Earth’s surface)
We need a more general expression for U based on Eqn. I.A.1
As before, we define potential energy in terms of the work done by the force associated with the potential energy:
This leads to the following definition of gravitational potential energy:
U = –GmEm/r (I.C.1)

5. U increases (becomes less negative) with increasing r
I.C. Gravitational Potential Energy
5. For distances greater than the Earth’s radius:
This definition of U means that U = 0 when r = 
What about a hollow sphere? U RE r RE
U increases (becomes less negative) with increasing r
–GmEm / RE

6. I.D. Escape Speed 1. Consider a projectile fired from a cannon
If gravitational force is the only force that does work, then mechanical energy is conserved: K1 + U1 = K2 + U2
For rocket to just barely escape to large values of r (say r = ), K2 = 0
At r = , U2 = 0 as well
Therefore K1 + U1 = 0, where K1 and U1 are measured at the launch point (ground):
1/2mv2 + (–GmEm/ RE) = 0
 vesc = [(2GmE)/ RE]1/2 = (2gRE)1/2 (I.D.1)
= km/s ~ 24,000 mph

7. I.E. Satellite Motion c) From Newton’s 2nd Law:
GmEm / r2 = marad = mv2 / r, so
vc = (GmE /r)1/2 = vesc/√2 (I.E.1)
“Circular Orbit Speed”
In terms of the period T of the orbit:
(II.E.2)

8. I.E. Satellite Motion
Example: What is the orbital period for “Low Earth Orbit?” For LEO, assume that r ~ RE (good assumption since r is typically RE km) and circular orbit:
T ~ 2pRE3/2/√(GME) ~ 90 minutes.
note: vc(LEO) ~ 8 km/s = 17,000 mph

9. I.E. Satellite Motion
Example: What is the orbital radius for a geosynchronus orbit?
Ts/TLEO = 24 hrs/1.5 hrs = r3/2/RE 3/2;
rs/RE = 162/3 = 6.3, or rs ~ 6RE ~ 40,000 km.
vs /vc(LEO)= (RE/rS)1/2 = 0.4, or
vs = 0.44vc(LEO) ~ 3 km/s