Gravitational potential energy and escape velocity

How much energy does it take to send a spaceship far from the Earth? Let's find out. We'll use a simple model for a spaceship: a big bullet of mass M = 1 kg shot out of a big gun, like this one:

Illustration from Jules Verne's book From the Earth to the Moon.

Worksheet for this exercise

The bullet starts at the Earth's surface, a distance R1 = 6.37 x 10⁶ meters from the center of the Earth, and then flies to a final distance of R2 = 2 x 6.37 x 10⁶ meters, where it stops momentarily.

How much gravitational potential energy has it gained?
What was the bullet's kinetic energy when it left the gun?
What was the bullet's speed when it left the gun?

The bullet starts at the Earth's surface, a distance R1 = 6.37 x 10⁶ meters from the center of the Earth, and then flies to a final distance of R2 = 11 x 6.37 x 10⁶ meters, where it stops momentarily.

How much gravitational potential energy has it gained?
What was the bullet's kinetic energy when it left the gun?
What was the bullet's speed when it left the gun?

The bullet starts at the Earth's surface, a distance R1 = 6.37 x 10⁶ meters from the center of the Earth, and then flies to a final distance of R2 = 101 x 6.37 x 10⁶ meters, where it stops momentarily.

How much gravitational potential energy has it gained?
What was the bullet's kinetic energy when it left the gun?
What was the bullet's speed when it left the gun?

(Yes, we are ignoring the effect of the Moon's gravitational force ...)

The bullet starts at the Earth's surface, a distance R1 = 6.37 x 10⁶ meters from the center of the Earth, and then flies to a final distance of R2 = 1,001 x 6.37 x 10⁶ meters, where it stops momentarily.

How much gravitational potential energy has it gained?
What was the bullet's kinetic energy when it left the gun?
What was the bullet's speed when it left the gun?

The bullet starts at the Earth's surface, a distance R1 = 6.37 x 10⁶ meters from the center of the Earth, and then flies to a final distance of R2 = 10,001 x 6.37 x 10⁶ meters, where it stops momentarily.

How much gravitational potential energy has it gained?
What was the bullet's kinetic energy when it left the gun?
What was the bullet's speed when it left the gun?

(Yes, we are ignoring the effect of the Sun's gravitational force, and that of the other planets ....)

The bullet starts at the Earth's surface, a distance R1 = 6.37 x 10⁶ meters from the center of the Earth, and then flies to a final distance of R2 = 100,001 x 6.37 x 10⁶ meters, where it stops momentarily.

How much gravitational potential energy has it gained?
What was the bullet's kinetic energy when it left the gun?
What was the bullet's speed when it left the gun?

(Yes, we are ignoring the effect of the Sun's gravitational force, and that of the other planets ....)

The bullet starts at the Earth's surface, a distance R1 = 6.37 x 10⁶ meters from the center of the Earth, and then flies to a final distance of R2 = 1,000,001 x 6.37 x 10⁶ meters, where it stops momentarily.

How much gravitational potential energy has it gained?
What was the bullet's kinetic energy when it left the gun?
What was the bullet's speed when it left the gun?

(Yes, we are ignoring the effect of the Sun's gravitational force, and that of the other planets ....)

Summary of the results



 R1 (km)             R2 (km)        gained GPE (J)    initial speed (m/s)
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6.37x10^6        2 x 6.37x10^6     31.3     million         7,913 

                11 x 6.37x10^6     56.9     million        10,670

               101 x 6.37x10^6     62.0     million        11,135

             1,001 x 6.37x10^6     62.55    million        11,185

            10,001 x 6.37x10^6     62.61    million        11,190

           100,001 x 6.37x10^6     62.616   million        11,190.7

         1,000,001 x 6.37x10^6     62.61626 million        11,190.73

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There's a clear pattern: at some point, the bullet gains a VERY large extra distance for just a small increase in initial velocity. Suppose that we make the final distance R2 = infinity. We can use the conservation of energy to find the speed required to keep the bullet going forever:

If we ask the bullet to reach a final distance R_f = ∞ with a final speed v_f = 0, then

That allows us to define a special initial speed, the escape velocity from the Earth:

We can finish our table now ....



 R1 (km)             R2 (km)        gained GPE (J)    initial speed (m/s)
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           100,001 x 6.37x10^6     62.616   million        11,190.7

         1,000,001 x 6.37x10^6     62.61626 million        11,190.73

               infinity            62.61633 million        11,190.74
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