Download Chapter 7 Problem 56 pdf

Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Chapter 7 P56 Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN
#) 978-049511374-3
THE PROBLEM STATEMENT
Ch 7 P56 Show that the escape speed from the surface of a planet of uniform density is
directly proportional to the radius of the planet.
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Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 7 P56 Show that the escape speed from the surface of a planet of uniform density is
directly proportional to the radius of the planet.
Basic Solution (including BRAINSTORMING-Definitions, concepts , principles and Discussion
Given:
Find: Show that vescape at surface, R, is proportional to density, .
KE(r) + PE(r) = KE(r=infinity) +
PE(r=infinity)
(1)
E (J)
vinfinity >0
2
PE = -GMm/r where M is the mass of
the planet (parent – this includes stars)
and m is the mass of the satellite
which includes planets, moons, even
smaller stars around more massive
stars. M >>m.
Etotal >0 1/2mv - GMm/r < 0
1/2mv - GMm/r = 0
Etotal=0
r top
Etotal <01/2mv - GMm/r > 0
vinfinity >0
2
r (m)
2
PEgravity=-GMm/r
To escape the body of mass m must
just reach r= infinity with no energy
left. Since PE = 0 when r =infinity,
but KE = 0 must also be true. So v=0
at r=infinity.
½ mvescape(r)2 - GMm/r
= 0 [v=0] - 0.
So, mve(r)2 = 2GMm/r
(2)
cancel m gives
ve(r)2 = 2GM/r
Hence
ve(r) = sqrt(2GM/r)
(3)
This is the v that the body of mass m must have at a distance r from the center of, and outside of
M.
Some definitions:
Volume of sphere = 3/4*R3
Density  = mass/volume M/V = M/(3/4*R3) = 3 MR3 /4 .
So M = V = R3/3),
Now ve(r) = sqrt(2GM/r) becomes
ve(r) = sqrt(2G(R3 /3r) = sqrt(8GR3 )/3r)
If one escaping from the surface of the planet, then r = R, so,
ve(R) = sqrt(8GR3 /3R) = sqrt(8GR2 /3) = Rsqrt(8G)
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(3)
(4)