precisioninfo.com - Dr Bryan Hall - 2004

Motion Under Gravity & the Escape Velocity of a Rocket

• Universal Law of Gravitational Attraction
  • Calculating acceleration due to gravity (g)
• Force Acting on the Rocket
• Rocket velocity at a distance x from the centre of the earth
• How high the rocket will rise?
• Rocket escape velocity
• Notes

Universal Law of Gravitational Attraction

The force between any two particles having masses m₁ and m₂ separated by a distance r is an attraction acting along the line joining the particles and has magnitude

where G is a universal constant having the same value for all pairs of particles. The present accepted value of G is (Physics - Resnick & Halliday):

It is a fact of spherical geometry that the gravitational force acting between any two homogeneous spheres having masses m₁ and m₂ separated by a distance r is exactly as stated for the particulate case.

Exercise

Check that these units of G provide correct units for the force F_G

Where we are considering a rocket escaping from the earth we write:

Where m_r denotes the mass of the rocket and m_e that of the earth

Calculating acceleration due to gravity (g)

The radius of the earth is stated as 6378.1 kilometres (Google Calculator) and the mass of the earth is reported as 5.9742 × 10²⁴ kilograms (Google Calculator). At the surface of the earth we therefore obtain:

This is the origin of g and note the units of g are exactly as required.

Further observe g has a representation as where R is the radius of the earth

Note this result does not hold as objects move further away from the surface of the earth.

Force Acting on the Rocket

The force acting on the rocket at all points above the surface of the earth is

Hence, at all points above the surface of the earth and neglecting friction, the single force acting on the rocket is F_G (exactly as shown). From this we can get the rockets acceleration (the rocket is in fact being slowed by gravity so it is deceleration). Writing a_r for the acceleration of the rocket and having regard to the consistent use of signs we obtain:

Exercise:

Apply the representation to formulate an equivalent expression for a_r in terms of g, R and the distance r between the centre of the earth and that of the rocket. The following diagram shows the relationship between R and r. (This problem is substantially solved below where g is inserted into the final velocity expression.)

Exercise:

What would be determined on dividing F_G by m_e instead of m_r. The answer is not as obvious as you may at first think - particularly in the case where two spheres of equivalent mass are under consideration.

Rocket velocity at a distance x from the centre of the earth

Considering the case only in so far as the rocket is not returning to the earth and changing the separation distance from r to x we can write:

Þ

Þ

Þ

The initial conditions suggest that, on the surface of the earth (R = 6378.1 km), we have:

Þ

Substituting this in we obtain:

Þ

Þ

Þ (on recalling the representation )

How high the rocket will rise?

The rocket will rise for so long as the velocity remains positive; hence we obtain the solution to this problem by setting v = 0

Þ

Þ

Þ

Rocket escape velocity

We obtain the escape velocity by requiring in limiting case x ® ¥ that the expression

remain positive. That is we require

Þ

Þ

On substituting the values for g and R we obtain the escape velocity v_o as:

Þ

Converting from metres per second to kilometres per hour yields approximately 40,000 kilometres per hour.

Notes:

• The gravitational forces acting on the rocket by bodies other than the earth have been ignored. There will be an appreciable gravitational attraction between the moon, other planets and the sun at substantial distances from the earth.
• The frictional effects of the atmosphere and the effects of rotation of the earth about its own axis have been set aside.
• The force to launch a rocket is provided by the thrust of its engines. The rocket takes a finite time to reach its escape velocity. While the rocket is climbing it is using fuel and therefore the overall mass is decreasing. Consequently, if the engines were to push with constant upward thrust, the acceleration of the rocket due to the engines would increase since the overall mass is decreasing.

References

• Physics. D Halliday and R Resnick (Third Edition). Published by John Wiley & Sons (1978).
• Google Calculator: http://www.google.com/help/calculator.html