Gravitational Force - Physics with Dr. Winters - Home
Transcript of Gravitational Force - Physics with Dr. Winters - Home
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Gravitational Force
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Gravitational Force
Newton’s Law of Gravitation:
G = 6.67 x 10-11 Nm2/kg2
= gravitational constant
rr
mGmFG
ˆ2
21F
-F
r
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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
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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
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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.
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Gravitational Potential Energy
Gravitational potential energy is defined as zero (0) at infinite distance (r∞).
Total energy: ET = K + UG
r
mGmUG
21
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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!
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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
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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.
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Orbit Problem
a) Calculate the radius:
m
ssm
vTr
T
rv
61083.3
2
)7080)(/3400(
2
2
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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
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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
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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
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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
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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
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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.
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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(