Key for AST3018 HMW 2
31 Points
Leah Simon
October 1, 2009
2.1 (3 pts)
Using the equation:
v
esc
=
2
GM
r
(1)
and the values G = 6.67
×
10
−
11
m
3
/kg/s
2
, M = M
Sun
= 1.989
×
10
30
kg, and
r = Earth’s Orbit = AU = 1.496
×
10
11
M we get
v
esc
= 4.21
×
10
4
m/s.
Using the equation:
v
orbit
=
GM
r
(2)
We find
v
orbit
= 2.98
×
10
4
m/s. Therefore,
v
esc
/v
orbit
=
2
.
2.2 (3 pts)
Using the equation:
F
=
GMm
r
2
=
mg
(3)
we find there is a relationship such that
M
Earth
R
Earth
2
=
a
M
obj
R
obj
2
(4)
Where
a
is a constant relating g
object
to g
Earth
. Therefore, using values taken from the book, we find
a.)
g
Moon
= 0.17
g
Earth
b.)
g
Sun
= 27.9
g
Earth
and c.)
g
Jupiter
= 2.54
g
Earth
2.3 (3pts)
Use the equation W = mg where W is the weight in Newtons, m is the mass in kg and g is
the gravitational acceleration on Earth,
a) on Earth W = 75 kg * 9.8 m s
2
= 735 N
b) on the Moon, recall from previous question
g
Moon
= 0.17
g
Earth
so
W = 75 kg * 0.17 * 9.8 m s
2
= 121 N
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c) in the space shuttle at an altitude of 400 km, g = GM/R
2
, where M = M
Earth
,
R = R
Earth
+ 400 km = 6.778 x 10
6
m, then
g
shuttle
= 8.677 m s
2
and W = 75 kg * 8.677 m s
2
= 651 N
2.4 (3 pts)
This is a large problem, but quite simple to solve with the right equations. First we start
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 Fall '08
 Lada
 pts

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