Solution for Longanswer Homework 24 Gravitation Problems
Solution to Longanswer Homework Problem 24.1()
Problem:
We often use
Δ
U
=
mg
Δ
y
to approximate the change in gravitational potential energy of an object
close to the surface of the earth, with
g
e
= 9
.
8
m
s
2
. Similarly, we can approximate the change in gravitational
potential energy of an object close to the surface of the sun as
Δ
U
s
=
mg
s
Δ
y
where
m
is the mass of the object,
g
s
is the gravitational acceleration at the surface of the sun, and
Δ
y
is the
height above the surface.
(a)Calculate
g
s
at the surface of the sun.
(b)Will the gravitational potential approximation given above for the sun be accurate over a larger or
smaller range of values of
Δ
y
than that for the earth? Justify your answer (do the math).
Solution to Part (a)
The gravitational force between two objects is given by
F
g
=

Gm
1
m
2
r
2
1
,
2
(the negative sign means the force is attractive) and by Newton II, we know that
F
=
ma
So taking
m
1
=
M
s
and
m
2
=
m
, and
a
=
g
s
at the surface of the sun (
r
1
,
2
=
R
s
), we have
mg
s
=
GM
s
m
R
2
s
g
s
=
GM
s
R
2
s
=
parenleftBig
6
.
673
×
10
−
11 Nm
2
kg
2
parenrightBig
(
2
.
0
×
10
30
kg
)
(7
.
0
×
10
8
m)
2
= 272
.
4
m
s
2
Grading Key: Part (a) 2 Points
Solution to Part (b)
Gravitational potential is given by
U
(
vector
r
) =

GMm
r
So the amount that gravitational potential changes over a (radial) distance
Δ
r
is
Δ
U
=
U
f

U
i
=

GMm
parenleftbigg
1
r
f

1
r
i
parenrightbigg
=
GMm
parenleftbigg
1
r
i

1
r
f
parenrightbigg
We are working this out for a distance
Δ
r
above the surface of the sun, so if we take
r
i
=
R
s
and
r
f
=
R
s
+Δ
r
,
we have
Δ
U
=
GMm
parenleftbigg
1
R
s

1
R
s
+ Δ
r
parenrightbigg
1
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Now we have to get a little creative (or notice that this derivation is in the book). Taking the second term in
parentheses, we can divide both the numerator and denominator by
R
s
, giving
1
R
s
+ Δ
r
=
1
R
s
1 +
Δ
r
R
s
=
1
R
s
parenleftbigg
1 +
Δ
r
R
s
parenrightbigg
−
1
Now the binomial theorem tells us that for small
x
, we have
(1 +
x
)
n
≈
1 +
nx
So as long as
Δ
r
is very small relative to
R
s
, we have
1
R
s
+ Δ
r
=
1
R
s
parenleftbigg
1 +
Δ
r
R
s
parenrightbigg
−
1
≈
1
R
s
parenleftbigg
1

Δ
r
R
s
parenrightbigg
And we can approximate
Δ
U
as
Δ
U
≈
GMm
bracketleftbigg
1
R
s

1
R
s
parenleftbigg
1

Δ
r
R
s
parenrightbiggbracketrightbigg
=
GMm
Δ
r
R
2
s
And so
Δ
U
≈
m
parenleftbigg
GM
R
2
s
parenrightbigg
(Δ
r
) =
mg
s
Δ
r
All this is wonderful, but the question asked us whether the approximation
Δ
U
=
mg
s
Δ
r
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 Spring '08
 Stewart
 Physics, Energy, Mass, Potential Energy, Work, General Relativity

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