(This project involves solving a differential equation by separation of variables.)
When an object is thrown upwards, we know by experience that it falls back to earth. How is it that we can launch rockets into space? We will investigate this question.
Let’s start with a mental experiment. When you throw a stone upwards it eventually falls back to earth. Does it make a difference how hard you throw the stone? Will it go higher if you throw it harder? Throwing the stone harder means a larger ‘initial velocity’. How hard would you have to throw to prevent the stone from falling back to earth?
Newton’s law of gravitation says that the force acting between two objects (the earth and our rocket) is proportional to their masses (M and m) and inversely proportional to the distance between their centers of gravity (d), that is,
F = ( G M m ) / d 2
where G (a constant) is there to make the units and values work out! (G = 6.67 · 10-11 N-m2 /kg 2 – check these units).
Newton’s law of motion says that F = m · a (mass times acceleration), so an object on the earth’s surface is subject to a force of F = m · g from the earth’s gravity, from which we conclude – if the two laws are consistent, that g = G M /R2 = (32 ft/s2 = 9.8 m/s 2 ) (R, earth’s radius). Verify the value of g using the fact that M = 5.98 ·1024 kg and R = 6378 km (watch units!, and do it for both systems).
When we launch a rocket, the distance between the rocket and the earth changes with time. When the rocket is h units (say miles) above the earth’s surface we can calculate the force acting on it using the law of gravitation:
F = (G M m) / (R+h)2= (G M /R2) m ( R2 / (R+h) 2 ) = g m R2 / (R+h)2 (*)
(the middle step involves multiplying and dividing by R2).
The differential equation:
Let’s assume, unrealistically, that the rocket engines shut off after launch, and so the rocket gets its acceleration from the initial thrust of the engines.
As mentioned, F = m · a, and a = dv/dt, so F = m · dv/dt . Combine this with equation (*) to find an expression for dv/dt (a differential equation). F and dv/dt are negative since the rocket is slowing after launch, so don’t forget to take this into account.
Changing variables: When we write dv/dt we are saying that v depends on t, ie, v=v(t).But in this case our differential equation shows that v also depends on h. Clearly h depends on t (h(t)) and dh/dt = v, so we must think of v as a composition: v=v(h(t)). Thus, by the chain rule, dv/dt = dv/dh · dh/dt. Use this to rewrite the differential equation so that we have an equation for dv/dh.
You will find the “escape velocity”, the initial velocity below which the rocket would fall back:
a) Verify the value of g from part 1 (and remember that most calculators can convert units).
b) Write the differential equation from part 2.
c) Solve the differential equation by separation of variables (it’s ok to leave the solution implicit: v2 = ···).
d) When h=0, v=v0 (initial velocity). Use this initial value to find the constant from (c).
e) If the rocket is going to “escape”, h →∞. Use this to let one of the terms in your solution → 0, and simplify the solution accordingly.
f) Using the data for the earth that you already have, calculate the “escape velocity” v0 , which you can determine by making sure that v0 is large enough so v >= 0. Give your answer in mi/hr and km/sec (careful with units).
g) The moon has a mass of 7.35 · 1022 kg and a radius of 1.738 · 106 m. Find the escape velocity for the moon.
Extra credit (use the library or the internet):
a) Find data for another planet and calculate its escape velocity.
b) Write a paragraph about Isaac Newton using several sources and citing references.
© David Rutschman, 2002