17.27 Getting into space

Before you read this, I suggest you read posts 17.19, 16.20 and 16.16. But post 17.19, and this one, involve some mathematics that you might wish to ignore – so you may prefer not to read this post after the first few paragraphs!

rocket-launch-67720_640

If a rocket is to be launched into space, it must take off with a minimum speed, perpendicular to the surface of the earth – the resulting velocity (post 17.4) is called the escape velocity. This speed is about 11 km.s-1 which is more than 30 times the speed of sound and much faster than a bullet leaves a gun (which is why bullets eventually fall to the ground, if shot into the air).

To get into space, the rocket must start with sufficient kinetic energy (post 16.21) to overcome the gravitational potential energy pulling it back to earth (post 16.21). Explaining the same thing an alternative way, the rocket must generate a force (post 16.13) to overcome the force of gravity (posts 16.16 and 16.17). As this force moves the rocket upwards, it does work (post 16.20).

The work done by the force in moving the rocket a very small distance, δr, is given (approximately) by

eqn 1 cropped

Here G is the universal gravitational constant (post 16.16), m is the mass of the rocket, M is the mass of the earth and r is the distance of the rocket from the centre of the earth. This equation is an approximation because F changes as r changes.

The total work done by the rocket is the sum of all these little increments of work written as

eqn 2 cropped

See post 17.19 to find out more about this notation. This result is an approximation because we’re adding together approximate vales for all the increments of work.

Now the mathematics starts! The expression for W becomes exact as δr becomes infinitesimally small (see post 17.19). So W is given exactly by

eqn 3 cropped

I have written this result as an indefinite integral (post 17.19). However, we want the rocket to move from the surface of the earth (where r = R, the radius of the earth) to a distance where the earth’s gravitational field is effectively zero (which is true when r is infinity). So we can replace this indefinite integral by the definite integral

eqn 4 cropped

To find out more about evaluating definite integrals, see post 17.19.

Where does this work come from? It is provided by the initial kinetic energy of the rocket (post 16.20). From the definition of potential energy (post 16.20), this gives the result that

eqn 5 cropped

where v is the speed of the rocket.

So, to escape the earth’s gravitational field and fly into space, the rocket must have a speed equal to

eqn 6 cropped

If we put the values of G (6.674 ×10-11 m3.kg-1.s-2), M (5.972 ×1024 kg) and R (6.371 × 106 m) into this equation, we get a result of 1.1 ×104 m.s-1 or 11 km.s-1.

 

Related posts

17.26 Rockets
17.16 Why do planes fly?

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