patel (kp9583) – HW12 – Mackie – (20211)
1
This printout should have 14 questions.
Multiplechoice questions may continue on
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beFore answering.
001
(part 1 oF 2) 10.0 points
The period oF the earth around the sun is
1 year and its distance is 150 million km From
the sun. An asteroid in a circular orbit around
the sun is at a distance 229 million km From
the sun.
What is the period oF the asteroid’s orbit?
Correct answer: 1
.
88632 year.
Explanation:
Let :
T
e
= 1 year
,
r
e
= 150 million km
,
and
r
a
= 229 million km
.
±rom Kepler’s laws,
T
2
e
r
3
e
=
T
2
a
r
3
a
T
a
=
p
r
a
r
e
P
3
/
2
T
e
=
p
229 million km
150 million km
P
3
/
2
(1 year)
=
1
.
88632 year
.
002
(part 2 oF 2) 10.0 points
What is the orbital velocity oF the asteroid?
Assume there are 365 days in one year.
Correct answer: 24187
.
6 m
/
s.
Explanation:
v
a
=
2
π r
a
T
a
=
2
π
(2
.
29
×
10
11
m)
1
.
88632 year
1 y
365 d
×
1 d
24 h
1 h
3600 s
=
24187
.
6 m
/
s
.
003
(part 1 oF 2) 10.0 points
Given:
G
= 6
.
67259
×
10
−
11
N m
2
/
kg
2
.
A 744 kg geosynchronous satellite or
bits a planet similar to Earth at a radius
1
.
95
×
10
5
km From the planet’s center. Its
angular speed at this radius is the same as the
rotational speed oF the Earth, and so they ap
pear stationary in the sky. That is, the period
oF the satellite is 24 h
.
What is the Force acting on this satellite?
Correct answer: 767
.
255 N.
Explanation:
Let :
G
= 6
.
67259
×
10
−
11
N m
2
/
kg
2
,
M
satellite
= 744 kg
,
and
T
= 86400 s
.
Solution:
v
=
2
π r
T
F
=
M
satellite
v
2
r
=
M
satellite
4
π
2
r
T
2
(1)
= (744 kg)
4
π
2
(1
.
95
×
10
8
m)
(86400 s)
2
= 767
.
255 N
.
004
(part 2 oF 2) 10.0 points
What is the mass oF this planet?
Correct answer: 5
.
87681
×
10
26
kg.
Explanation:
Using the general gravitation law and Eq.
1, we have
F
=
G
M
satellite
M
planet
r
2
=
M
satellite
4
π
2
r
T
2
G
M
satellite
M
planet
r
2
=
M
satellite
4
π
2
r
T
2
,
so
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2
M
planet
=
4
π
2
GT
2
r
3
=
4
π
2
(6
.
67259
×
10
−
11
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 Fall '09
 MIHALISIN
 Physics, Correct Answer, Pressure measurement

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