karna (pk4534) – HW 8 – coker – (58245)
1
This
printout
should
have
17
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
This assignment involves the last chapter
that will be covered on upcoming Quiz 2.
001
10.0 points
Which of the objects
a. a book
b. the nearest star
c. the Sun
d. a distant galaxy
exert(s) a gravitational force on you?
1.
c
2.
b
3.
a and b
4.
b, c and d
5.
Another combination
6.
c and d
7.
a
8.
a and d
9.
a, b, c and d
correct
10.
a, b and c
Explanation:
All objects exert a gravitational force on
you, no matter how far away they are.
002
10.0 points
Planet X has nine times the diameter and
two times the mass of the earth.
What is the ratio
g
X
:
g
e
of gravitational
acceleration at the surface of planet X to the
gravitational acceleration at the surface of the
Earth?
1.
g
x
g
e
=
8
9
2.
g
x
g
e
=
3
4
3.
g
x
g
e
=
4
9
4.
g
x
g
e
=
9
64
5.
g
x
g
e
=
1
12
6.
g
x
g
e
=
2
81
correct
7.
g
x
g
e
=
1
7
8.
g
x
g
e
=
1
4
9.
g
x
g
e
=
1
2
10.
g
x
g
e
=
6
49
Explanation:
Let :
M
x
= 2
M
e
,
and
R
x
= 9
R
e
.
mg
e
=
GM
e
m
R
2
e
g
e
=
GM
e
R
2
e
.
Similarly,
g
x
=
GM
x
R
2
x
.
The ratio is
g
x
g
e
=
M
x
R
2
e
M
e
R
2
x
=
2
M
e
R
2
e
M
e
(9
R
e
)
2
=
2
81
.
keywords:
003
10.0 points
The
mass
of
a
certain
neutron
star
is
9
×
10
30
kg (4
.
5 solar masses) and its radius
is 3500 m.
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karna (pk4534) – HW 8 – coker – (58245)
2
What is the acceleration of gravity at the
surface of this condensed, burnedout star?
The value of the universal gravitational con
stant is 6
.
67
×
10
−
11
N
·
m
2
/
kg
2
.
Correct answer: 4
.
90041
×
10
13
m
/
s
2
.
Explanation:
Let :
M
= 9
×
10
30
kg
,
r
= 3500 m
,
and
G
= 6
.
67
×
10
−
11
N
·
m
2
/
kg
2
.
The gravitational acceleration is
g
=
GM
r
2
=
(
6
.
67
×
10
−
11
N
·
m
2
/
kg
2
)
(9
×
10
30
kg)
(3500 m)
2
=
4
.
90041
×
10
13
m
/
s
2
,
several hundreds of billion times
g
on earth.
004
(part 1 of 2) 10.0 points
You weigh 300 lb.
How much would you weigh if you were
standing on a mountain 200 km tall (equiva
lent to standing still at about the altitude of a
space shuttle orbit)? 4.45 N = 1 lb, the grav
itational constant is 6
.
67
×
10
−
11
N
·
m
2
/
kg
2
,
the radius of the Earth is 6
.
4
×
10
6
m, and the
mass of the Earth is 6
×
10
24
kg.
Correct answer: 1251
.
54 N.
Explanation:
Let :
W
= 300 lb
,
h
orbit
= 200 km = 2
×
10
5
m
,
r
E
= 6
.
4
×
10
6
m
,
and
G
= 6
.
67
×
10
−
11
N
·
m
2
/
kg
2
.
The gravitational acceleration at an alti
tude of 200 km is
g
′
=
GM
E
r
′
2
=
GM
E
(
r
E
+
h
orbit
)
2
=
(6
.
67
×
10
−
11
N
·
m
2
/
kg
2
) (6
×
10
24
kg)
(6
.
4
×
10
6
m + 2
×
10
5
m)
2
= 9
.
18733 m
/
s
2
.
Your weight is
W
= 300 lb
×
4
.
45 N
lb
= 1335 N
,
so your mass is
m
=
W
g
=
1335 N
9
.
8 m
/
s
2
= 136
.
224 kg
.
At an altitude of 200 km, your weight would
be
W
′
=
mg
′
= (136
.
224 kg) (9
.
18733 m
/
s
2
)
=
1251
.
54 N
.
005
(part 2 of 2) 10.0 points
How much does this differ from your weight
on the surface of the Earth?
Correct answer: 18
.
7553 lb.
Explanation:
W  W
′
= 1335 N

1251
.
54 N
= 83
.
461 N
×
1 lb
4
.
45 N
=
18
.
7553 lb
,
which would be easily detected on a bathroom
scale.
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 Spring '08
 Turner
 Physics, Energy, Mass, General Relativity, Correct Answer

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