Transcript Newton derives Kepler`s laws

CAPSTONE Lecture 4
Laws of planet motion
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1. What are the units of G? (Set F=ma=-GmM/r2, solve
for G, and work out the resulting units.)
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2. Calculate the orbital velocity of Earth around the Sun.
Use the circumference of the (assumed) circular orbit
and the time it takes for one orbit to get the velocity. This
should give the same answer as the equation
vE=(GMS/r)1/2, where vE is the unknown velocity of Earth
in its orbit, G is the Gravitational constant, Ms is the mass
of the Sun and r is the Earth-Sun distance.
(one astronomical unit, r=1.5 x 1013 cm.)
Calculate the orbital velocity of Jupiter around the Sun.
Calculate the orbital velocity of the Moon around the
Earth
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3. a) What is the kinetic energy of a 55 kg runner while
running the 100 yard dash in 10 seconds? Use her
average velocity.
b) What is the kinetic energy of Jupiter in motion around
the Sun?
4. a) What is the potential energy of Jupiter in orbit around
the Sun? (Ignore the other planets and nearby stars.)
b) Verify that K=-U/2 for an object in orbit around another
body.(the Virial theorem.)
Do this using the numbers above, and analytically.
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Kepler’s Law
• P2=R3, P=period, R=average orbit
radius.
• Consider a body with M1, v1 at distance
r from M2, v2. R=r1+r2. r1 and r2
distances from the point about which
both objects seem to move (center of
mass).
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Center of Mass
• (M2v22)/r2=(M1v12/r1). (Huygens)
• v22/v12=M1r2/M2r1=(2r2/P2)2/ (r1/P1)2,
since, P1=2r1/v1, P2=2r2/v2
(1)
(2)
(3)
But, note, P1=P2 (both complete one round at the same time)
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v22/v12=M1r2/M2r1=(2r2/P2)2/ (r1/P1)2
From ( 2), M1r2/M2r1=r22/r12 so M1r1=M2r2
(4)
This is the definition of center of mass. The more massive object
is closer to the center of mass.
From eq. 3 and 4, r1/r2=M2/M1=v1/v2
The more massive object is closer to the center of mass and
moves slowest.
For newly discovered, extra-solar planets, it is the very tiny vel.
of the massive star that is seen (<10 m/sec)
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Final step
• The force acting on any one of the two bodies can be expressed
in terms of the overall system, or the center of mass system.
The consequences are the same.
• Pick the second body, there are two expressions for force that
are equal.
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F=-GM1M2/(r1+r2)2=-M2v22/r2.
(6)
F=…………………..=[-M2v22/r2][42r2/ 42r2]
F=……………………=[M2 42r2][v22/ 42r22]
(7)
But, the last term is just [1/P2], where P=2r2/v2
GM1M2/R2 = M2r24 / P2 (since R=r1+r2)
GM1M2/R2 = M2r24 / P2 (since R=r1+r2)
• GM1/R2 = r24 / P2
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Conclusion
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Now, rewrite some terms:
R=r1+r2=r2(1+r1/r2), or r2=R/ (1+r1/r2)
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M=M1+M2=M1(1+M2/M1) or M1=M / (1+M2/M1)
Form the ratio M/R
M/R =[ M1(1+M2/M1)] / r2(1+r1/r2)]
From the equation 4), r1/r2= M2/M1
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M/R = [ M1(1+M2/M1)] / r2(1+M2/M1)] = M1/r2
M/R = [ M1(1+M2/M1)] / r2(1+M2/M1)] = M1/r2
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Equation 8 tells us that
GM1/R2 = r24 / P2, so G(M1/r2)/R2 = 4 / P2
G(M/R)/RR2 = 4 / P2
GM/R3 = 4 / P2
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If we talk about the Earth/Sun system, we
know R=1AU, P=1 year, M = 1 solar mass.
Convert G (cm3/g-sec2) to
G (AU3/M(solar)-yr2), then 42/G=1 and
R(AU)3=P(yr)2.
Newton derived from first principles the same
empirical law as Kepler for planetary motion.
QED
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42/GM=1 in units of AU, Mo, yr.
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P(s)2={4/[GM(Mo)]}R(AU)3, where Mo=1 solar mass
If Kepler finds P(yr)2=R(AU)3, the true form must be
P(yr)2=R(AU)3/M(Mo)
Therefore, 42/GM = 1, M=1, 42/G=1
or G=4 2 in solar system units
G=6.67 x 10-8 cm3/g-sec2, 1cm=1AU/1.5x1013 cm,
1s=1yr/ x107s, 1g=1Mo/ 2x1033.
G=[6.67 x 10-8 x(1AU/1.5x1013)3]/[(1
Mo/2x1033)(1yr/x107)2)]
G=[6.67x22/(1.5)3]x[10-8x10-39 x1033x1014][AU3/Moyr2]
G=[(13.34/3.375)] x 2 x 10-47 x 1047 [AU3/Mo-yr2]
G=3.95 2
QED
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