2.3.3 In Class or Homework Exercise
1. Using Newton's Law of Universal Gravitation,
a. Show that for any planet in a circular orbit around the sun, the ratio
R3 / T 2 is a constant
We know that Newton’s equation for universal gravitation can be used
to calculate the centripetal force required to keep a planet in orbit:
Fc  mac
Gm p ms
r2
Gm p ms
 mac
 mp
r2
Gms
 v2
r
v2
r
Since we are assuming a circular orbit, the speed of the planet is given
2 r
by v 
.
T
Gms  2 r 


r
 T 
Gms 4 2 r 2

r
T2
Gms r 3

4 2 T 2
2
Since G and 4 2 are constant values, and ms is constant as long as we
are comparing planets that are orbiting the sun, then the ratio
r3
T2
Is a constant.
b. Using the earth’s orbital radius and period, find the value of this
constant.
RES  1.50 1011 m
TE  365.25days  3.16 107 s
r 3 (1.50 1011 )3

T 2 (3.16 107 ) 2
 3.38 1018 m3 / s 2
UNIT 2 2D Motion
RRHS PHYSICS
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c. Calculate the mass of the sun.
From part a),
Gms r 3

4 2 T 2
(6.67 1011 )ms
 3.38 1018
2
4
ms  2.00 1030 kg
Which is very close to our accepted value of 1.99 1030 kg .
2. Use Kepler's third law and the period of the moon (27.4 days) to find the
height of a geosynchronous satellite.
The two objects that we will compare are the moon and the geosynchronous
satellite:
rme  3.85 108 m
Tm  27.4d
Tg  1d
rge  ?
3
rge3
rme

Tm2 Tg2
r  Re  h
3
(3.85 108 )3 rge

(27.4) 2
(1) 2
4.24 107  6.38 106  h
h  3.6 107 m
rge3  7.60 1022
rge  4.24 107 m
This is the same answer that was obtained in the previous section.
3. Jupiter is 5.2 times farther than Earth is from the sun. Find Jupiter's orbital
period in Earth years.
TJ  ?
TE  1 year
RES  R
RJS  5.2 R
UNIT 2 2D Motion
RRHS PHYSICS
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3
3
RJS
RES

TJ2
TE2
(5.2 R)3 ( R)3

TJ 2
(1) 2
140.6 R 3 R 3

TJ 2
1
TJ  11.9 years
4. Uranus requires 84 years to circle the sun. Find Uranus' orbit as a multiple of
Earth's orbital radius.
TU  84 years
TE  1 year
RES  R
RUS  ?
3
3
RUS
RES

TU2
TE2
RUS 3 RES 3

(84) 2 (1) 2
RUS  19.2 RES
5. A satellite is placed in an orbit with a radius that is half the radius of the
moon's orbit. Find its period in units of the period of the moon.
Ts  ?
RsE  12 R
RmE  R
3
3
RmE
RsE

Tm2
Ts2
R 3 ( 12 R )3

Tm 2
Ts 2
R 3 18 R 3
 2
Tm 2
Ts
Ts 2  18 Tm 2
Ts  0.35Tm
UNIT 2 2D Motion
RRHS PHYSICS
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6. A friend tells you that because of Kepler’s Second Law, Mars will sweep out
an equal area in its orbit as the earth will in the same amount of time. Is he
right?
No. Kepler’s Second law applies to a single planet, indicating that it will sweep
out equal areas in equal times. It cannot be used to compare planets.
UNIT 2 2D Motion
RRHS PHYSICS
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