lavallo (jhl936) – Homework 11 – florin – (58140)
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001
10.0 points
A satellite circles planet Roton every 4
.
9 h in
an orbit having a radius of 1
.
7
×
10
7
m.
If the radius of Roton is 1
.
292
×
10
7
m, what
is the magnitude of the freefall acceleration
on the surface of Roton?
Correct answer: 3
.
73409 m
/
s
2
.
Explanation:
Basic Concepts:
Newton’s law of gravi
tation
F
g
=
G
m
1
m
2
r
2
.
Kepler’s third law
T
2
=
parenleftbigg
4
π
2
G M
parenrightbigg
r
3
.
The freefall acceleration
a
on the surface of
the planet is the acceleration which a body in
free fall will feel due to gravity
F
g
=
G
M m
R
2
=
m a ,
where
M
is the mass of planet Roton.
This
acceleration
a
is
a
=
G
M
R
2
,
(1)
the number which is
g
on Earth. Here, how
ever, the mass
M
is unknown, so we try to
find this from the information given about the
satellite. Use Kepler’s third law for the period
of the orbit
T
2
=
parenleftbigg
4
π
2
G M
parenrightbigg
r
3
.
(2)
By multiplying both sides with
R
2
and com
paring to equation (1), we can identify our
a
in the right hand side
T
2
R
2
=
parenleftbigg
4
π
2
a
parenrightbigg
r
3
.
If we solve for
a
, we obtain
a
=
parenleftbigg
4
π
2
T
2
R
2
parenrightbigg
r
3
= 3
.
73409 m
/
s
2
which is our answer. Although identifying
a
in this way is a “quick” way of solving the
problem, we could just as well have calculated
the planet mass
M
explicitly from equation
(2) and inserted into equation (1).
002
10.0 points
Two planets A and B, where B has twice the
mass of A, orbit the Sun in elliptical orbits.
The semimajor axis of the elliptical orbit of
planet B is two times larger than the semi
major axis of the elliptical orbit of planet A.
What is the ratio of the orbital period of
planet B to that of planet A?
1.
T
B
T
A
=
√
8
correct
2.
T
B
T
A
=
1
8
3.
T
B
T
A
= 8
4.
T
B
T
A
=
1
4
5.
T
B
T
A
= 1
6.
T
B
T
A
= 2
7.
T
B
T
A
=
radicalbigg
1
2
8.
T
B
T
A
=
radicalbigg
1
8
9.
T
B
T
A
=
1
2
10.
T
B
T
A
=
√
2
Explanation:
Basic Concept:
Kepler’s Third Law is
T
2
=
parenleftbigg
4
π
2
G M
S
parenrightbigg
a
3
=
K
S
a
3
,
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lavallo (jhl936) – Homework 11 – florin – (58140)
2
where
K
S
=
4
π
2
G M
S
= 2
.
97
×
10
−
19
s
2
/
m
2
,
and where
a
is the semimajor axis of the
elliptical orbit of the planet (
a
=
r
the radius
of a planet in a circular orbit).
Solution:
According to Kepler’s third law,
the square of the orbital period is proportional
to the cube of the semimajor axis “
a
” of the
elliptical orbit. Therefore,
T
2
A
a
3
A
=
T
2
B
a
3
B
.
Therefore,
T
B
T
A
=
parenleftbigg
a
B
a
A
parenrightbigg
3
2
= 2
3
/
2
=
√
8
.
003
10.0 points
Given:
G
= 6
.
67259
×
10
−
11
N m
2
/
kg
2
Calculate the work required to move a
planet’s satellite of mass 571 kg from a cir
cular orbit of radius 2
R
to one of radius 3
R
,
where 8
.
8
×
10
6
m is the radius of the planet.
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 Spring '10
 Florin
 Kinetic Energy, Mass, Momentum, General Relativity

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