Gravitational Force
Gravitational Force
Newton’s Law of Gravitation:
G = 6.67 x 10-11 Nm2/kg2
= gravitational constant
rr
mGmFG
ˆ2
21F
-F
r
Gravitational Force
A uniform spherical shell of matter attracts a particle that is outside the shell as if all the shell’s mass were concentrated at its center
A uniform shell of matter exerts no net gravitational force on a particle located inside it.
rr
mGmFG
ˆ2
21r
FG = 0
Orbits
The centripetal force of an object orbiting around a planet or star (sun) is provided by the gravitational force:
2
2
2
2
211
r
Gm
r
v
r
mGmam
FF
c
Gc
r
m1
m2
Geosynchronous Orbit
Geosynchronous orbit:
orbit period (T) equals 1 day T
rv
2
2
2
32
2
2
22
2
21
2
1
2
21
2
1
4
4
2
TGm
r
r
Gm
T
r
r
mGm
r
T
r
m
r
mGm
r
vm
Can figure out r, m or T if you know the other two variables.
Gravitational Potential Energy
Gravitational potential energy is defined as zero (0) at infinite distance (r∞).
Total energy: ET = K + UG
r
mGmUG
21
Kepler’s 1st and 2nd Laws
1st: All planets move in elliptical orbits, with the sun at one focus.
2nd: A line that connects a planet to the sun sweeps out equal areas in equal times
Angular momentum is conserved!
Kepler’s 3rd Law
The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit.
For circular objects this follows directly from
Fc = FG
32
2 4r
GMT
Orbit Problem
In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars. Assume a circular orbit with a period of 7080s and orbital speed of 3400m/s. The mass of the GS is 930kg and the radius of Mars is 3.43 x 106 m.
a) Calculate the radius of the GS orbit.
Orbit Problem
a) Calculate the radius:
m
ssm
vTr
T
rv
61083.3
2
)7080)(/3400(
2
2
Orbit Problem
b) Calculate the mass of Mars.
T = 7080 s
v = 3400 m/s
mGS = 930 kg
rM = 3.43 x 106 m
rorbit = 3.83 x 106 m
Orbit Problem
b) Calculate the mass of Mars:
Set Fc = FG:
kg
msm
G
rvm
r
Gmv
r
mGm
r
vm
kg
mN
M
M
MGSGS
23
11
62
2
2
2
2
1064.6
1067.6
)1083.3()/3400(
2
2
Orbit Problem
c) Calculate the total mechanical energy of the GS in this orbit.
T = 7080 s v = 3400 m/s mGS = 930 kg mM = 6.64 x 1023 kg rM = 3.43 x 106 m rorbit = 3.83 x 106 m
Orbit Problem
c) Calculate the total mechanical energy of the GS in this orbit.
J
m
kgkgsmkg
r
mGmvm
UKE
kg
mN
MGSGS
GT
9
6
2311
2
21
2
21
1038.5
1083.3
)1064.6)(930)(1067.6()/3400)(930(
2
2
Orbit Problem
d) If the GS was to be placed in a lower circular orbit (closer to the surface of Mars), would the new orbital period of the GS be greater than or less than the given period? Justify your answer.
Answer: Less than. T2 is proportional to r3 so if r decreases
then T also decreases
Orbit Problem
e) In fact, the GS orbit was slightly elliptical with its closest approach to Mars at 3.71 x 105 m above the surface and its furthest distance at 4.36 x 105 m above the surface.
If the speed of the GS at closest
approach is 3.40 x 103 m/s, calculate the speed at the furthest point of the orbit.
Orbit Problem
e) Calculate the speed at the furthest point.
Angular momentum is conserved:
mmm
smmm
r
vrv
vrvr
r
vrm
r
vrm
II
f
ccf
ffcc
f
f
fGS
c
ccGS
ffcc
3
65
365
22
1034.3)1043.31036.4(
)/1040.3)(1043.31071.3(