This preview shows pages 1–4. Sign up to view the full content.
Solution for Practice Test 7 for Test 4
Solution to Practice Test Problem 7.1()
Problem:
(a)
NSF exam #
NSF point total
(b)
Grading Key: 45 Points
Grading Key: 25 Points
Total Points for Problem: 45 Points
Solution to Practice Test Problem 7.2()
Problem:
You are travelling across the solar system in your spaceship. As you are nearing a planet, the ship’s engine
sputters and dies, and you end up in circular orbit around the planet. You try several times to restart the engine, each
time altering your orbit, until fnally you decide to try something else. You start reading the ship’s log looking For ideas,
and you fnd the Following data For each orbit:
Height
Speed
1
×
10
7
m
6
×
10
4
m
/
s
1
.
5
×
10
7
m
5
×
10
4
m
/
s
2
×
10
7
m
4
×
10
4
m
/
s
2
.
5
×
10
7
m
3
.
5
×
10
4
m
/
s
3
×
10
7
m
3
×
10
4
m
/
s
Assume that “Height” measures the distance From the center oF the planet, and that each orbit is a stable, uniForm
circular orbit.
(a)Independantly oF the data, derive an equation relating
v
2
to
r
(that is, relating the square oF the orbital
speed to the radial distance From the planet’s center).
(b)Make a table oF
v
2
and
1
/r
values.
v
2
1
r
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document (c)Graph
v
2
vs.
1
/r
. Label your axes with units. Draw a bestft line.
(d)Calculate the slope oF your bestft line. Unclude units.
(e)Comparing your equation From part (a) to your graph in (c), give one reason related to the ship’s instru
ments why the data doesn’t go through the point (0,0). What type oF error does this suggest?
(F)Using the graph and techniques practiced in lab, decide which planet you are orbiting.
Planet
Mass
Jupiter
1
.
9
×
10
27
kg
Saturn
5
.
69
×
10
26
kg
Uranus
8
.
68
×
10
25
kg
(g)Are you confdent with your choice? Do a percent deviation analysis.
Solution to Part (a)
The only Force on the spacecraFt (
m
) is gravity due to the planet (
M
), so Newton II For the spacecraFt reads

Σ
V
F

=
ma
=
GMm
r
2
Assuming a circular orbit, the acceleration is centripetal, so
a
=
v
2
/r
,
m
p
v
2
r
P
=
GMm
r
2
v
2
=
GM
r
Grading Key: Part (a) 2 Points
1 point(s) : Correct answer.
1 point(s) : Reasonable derivation.
Solution to Part (b)
v
2
1
r
3
.
6
×
10
9
m
2
/
s
2
10
.
0
×
10

8
m

1
2
.
5
×
10
9
m
2
/
s
2
6
.
7
×
10

8
m

1
1
.
6
×
10
9
m
2
/
s
2
5
.
0
×
10

8
m

1
1
.
2
×
10
9
m
2
/
s
2
4
.
0
×
10

8
m

1
0
.
9
×
10
9
m
2
/
s
2
3
.
3
×
10

8
m

1
Grading Key: Part (b) 2 Points
1 point(s) (2 times) : Two middle values correct.
Solution to Part (c)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/04/2008 for the course PHYS 2054 taught by Professor Stewart during the Spring '08 term at Arkansas.
 Spring '08
 Stewart
 Physics

Click to edit the document details