Steven Charlton
Escape From Planet Earth
Expression for Escape Velocity
We will begin by deriving a general equation for the escape velocity from a body.
By the Work/Energy principle: Work  ΔEnergy
xm
0 ms-1
u ms-1
rm
F 
GMm
x2
v ms-1
Planet
∞m
The diagram above shows a body in various stages of its escape from a planet.
Newton's law of gravity tells us that the attractive force experienced between two bodies, of
mass M, and m, at a distance of x is given by:
F 
GMm
x2
G is the Universal Gravitational Constant, and is equal to 6.674 × 10-11 N m2 kg-2
On the surface of the planet, r meters from the centre, the object is launched away with a
speed v. If the object just manages to escape the gravitational pull of the body it will be just
able to reach an infinite distance and at that distance its velocity will become zero. The
velocity at which it is launched will then be the escape velocity.
The work done by the force can be calculated by integrating the force with respect to
distance between the start and end positions.
Work   Force dx

 
r
GMm
dx
x2

GMm 


 x r

GMm



GMm
r
GMm
r
The change in energy can be calculated by finding the difference between the start and end
velocities
ΔEnergy  Final  Initial
1
1
 m  0 2  mv 2
2
2
1
2
  mv
2
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Steven Charlton
Equating these we find
Work  ΔEnergy
GMm
1

  mv 2
r
2
Cancel the minus signs and an m on each sides gives:
GM 1 2
 v
r
2
If we solve this for v we find:
v 
2GM
r
And this gives the escape velocity in terms of the mass of the planet – M, and its radius – r.
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Steven Charlton
Escape from the Earth
To calculate the escape velocity of the Earth, we simply substitute the various values into the
expression bound above.
On the problem page the mass of the Earth is given as 5.9763 × 10 24 kg
The radius of the Earth is given as 6378 km = 6 378 000 m
If we work in SI units, Newtons, kilograms, meters, second, the units of velocity are then
meters per second.
Putting these in, along with the value for G noted above gives:
v 

2GM
r
2  6.674  10 11  5.9763  10 24
m s 1
6 378 000
 125 073 145 .8  m s 1
 11 183 .6  m s 1
 11.2 km s 1
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Steven Charlton
Escape from the Moon
Because the surface gravity of the moon is less than that of the Earth, the escape velocity of
the Moon will be less than that of the Earth.
We will now calculate it.
The problem page give the measurements of the Moon relative to the Earth. Rather than
calculating the moons measurements and than substituting in, it is simpler to substitute
directly in, and then find the Moon's escape velocity relative to the Earth. Then this can be
multiplied by what we know he Earth's is to calculate the Moon's escape velocity.
The Moon's mass is 0.0123 times the Earth's.
The Moon's radius is 0.273 times the Earth's.
If we let Me be the Earth's mass, and re be the Earth's radius and Mm, rm the moons, then we
can write:
M m  0.0123M e
and
rm  0.273re
Therefore:
v 
2GM
r

2G  0.0123M e
0.273re

0.0123 2GM e
0.273 re
 0.045054  
 0.21226  
2GM e
re
2GM e
re
The expression under the square root is simply the Earth's escape velocity, which we know
from above. Putting this value in we get:
v  0.21226   11 183 .6  m s 1
 2373 .8  m s 1
 2.37 km s 1
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Steven Charlton
Escape from Jupiter
Because Jupiter's surface gravity is greater than the Earth's, its escape velocity will be greater
than the Earth's.
As above we calculate the escape velocity relative to the Earth, and then evaluate that to get
avalue.
From the problem page:
Jupiter's radius is 11.209 times Earth's, and
Jupiter's mass is 317.8 times Earth's.
v 
2GM
r

2G  317 .8M e
11.209re

317 .8 2GM e
11.209 re
 28.352  
2GM e
 5.3246  
2GM e
re
re
 59 549 .147 m s 1
 59.5 km s 1
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Steven Charlton
Escape from the Sun
Because the Sun's surface gravity is much greater than the Earth's (and indeed Jupiter's), the
Sun's escape velocity will be much greater than Earth's (and Jupiter's).
From the problem page:
The Sun's radius is 109 times the Earth's, and
Its mass is 332946 times the Earth's
v 
2GM
r

2G  332946 M e
109re

332946 2GM e
109
re
 3054 .55  
 55.267  
2GM e
re
2GM e
re
 618095 .65  m s 1
 618 km s 1
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Steven Charlton
Planet Tiny
From the equation
v 
2GM
r
We can solve to find the radius given the mass and velocity. Solving the equation for r gives:
r 
2GM
v2
We know that Planet Tiny has the same mass as the Earth, therefore M = 5.9763 × 1024 kg.
We know the escape velocity, it is given as half the speed of light. The speed of light is
approximately 3.00 × 108 m s-1, so the escape velocity is 1.50 × 10^8 m s-1.
Putting this into the equation gives:
r 

2GM
v2
2  6.674  10 11  5.9763  10 24
1.5  10 
8 2
m
 0.035454 m
 3.55 cm
As the radius decrease the escape velocity will increase by the first equation above, and so
the radius of the planet tiny must be 3.55 cm or smaller to prevent the cannon ball from
escaping.
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Steven Charlton
Planet Heavy
From the equation
v 
2GM
r
We can solve to find the mass given the radius and velocity. Solving the equation for M
gives:
M 
rv 2
2G
We know that Planet Heavy has the same radius as the Earth, therefore r = 6 378 000 m.
We know the escape velocity, it is given as half the speed of light. The speed of light is
approximately 3.00 × 108 m s-1, so the escape velocity is 1.50 × 10^8 m s-1.
Putting this into the equation gives:

6 378 000  1.5  10 8
2  6.674  10 11
 1.07510   10 33 kg
M 

2
kg
 1.075  10 33 kg
As the mass increases the escape velocity increase, by the first equation above. This means
that the mass of Planet Heavy must be at least 1.075 × 10 33 kg.
The density of a body is defined as its mass divided by its volume. As both Planet Earth and
Planet Tiny have the same volume the ratio of their densities will simply be the ratio of their
masses.
Mass of Planet Heavy 1.075   10 33

Mass of the Earth
5.9763  10 24
 179 894 732
 179 900 000
The planet heavy must be at least 179.9 million times as dense as the Earth to prevent the
cannon ball from escaping.
The density of the earth can be calculated as mass over volume, where the volume is that of
4
a sphere, r 3
3
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Steven Charlton
Mass
Volume
Mass

4 3
r
3
5.9763  10 24

kg m- 3
4
3
 6 378 000 
3
5.9763  10 24

kg m- 3
1.08679129   10 21
 5 499 .082512  kg m- 3
Density 
 5 499 kg m- 3
The density of Planet Heavy is therefore
179 894 732  5 499.082512  kg m-3  9.892559727   1011 kg m-3
 9.893  1011 kg m- 3
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Steven Charlton
Escape from a Neutron Star
The problem page gives the mass of a typical Neutron star as 2 times the mass of the sun,
which is itself 332946 times the mass of the Earth. The radius of a Neutron star is given as
10 km, which is 10 000 m.
This means that the mass of a neutron star is:
M  2  332946  mass of the earth
 2  332946  5.9763  1024 kg
 3.97957  1030 kg
 3.9796  1030 kg
Putting this into the escape velocity equation gives:
v 

2GM
r
2  6.674  10 11  3.9796  10 30
m s 1
10 000
 5.31193   1016 m s 1
 230 476 257 .3  m s 1
 230 500 km s 1
This escape velocity is greater than that of either Planet Heavy, or Planet Tiny, and much
much greater than that of the Sun. This is more than 75% the speed of light.
Calculating the density of a Neutron star, in order to compare with Planet Heavy gives:
Mass
Volume
Mass

4 3
r
3
3.9796  10 30

kg m- 3
4
3
 10 000 
3
3.9796  10 30 kg

kg m- 3
12
4.188790205   10
 9.500524412 .  1017 kg m- 3
Density 
 9.501 .  1017 kg m- 3
We can see that the ratio of the densities of the Neutron star and Planet Heavy is:
Density of Neutron Star 9.500524412   1017

Density of Planet Heavy
9.892559727   1011
 960 370 .69
 960 000
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Steven Charlton
The Neutron star is almost one million times more dense than Planet Heavy.
To see how many times more dense the Neutron star is then Earth we can multiply the two
ratios we found: 960 370.69… × 179 894 732 = 1.900523003 × 1014. So we can see the
neutron star is about 1.90 × 1014 times more dense than the Earth itself.
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