Applications of calculus to the physical world

A rocket is fired vertically from the surface of the earth at a velocity of 5km per second. The acceleration of the rocket is given by a= -80000/x^2 kilometres per second, where x is the distance of the rocket from the centre of the earth.
If the radius of the earth is 6400km, find the rockets velocity when it is 3600 km above the earths surface.

Working out required

a = second derivative of distance with respect to time = (d^2 s)/dt^2 = -8000/x^2

integrating both sides, you have

first derivative of distance with respect to time = velocity = ds/dt = 8000/x + C1
The given "from the surface of the earth at a velocity of 5km per second." implies that when x = 6400 km, v = 5. We can solve for C1

8000/6400 + C1 = 5
5/4 + C1 = 5
C1 = 15/4

So now we have
velocity = v = ds/dt = 8000/x + 15/4

Therefore, at 3600 km above the earth's surface, the rocket is 3600 + 6400 km from the centre of the earth. x = 10000 km

v = 8000/10000 + 15/4 = 4/5 + 15/4 = 91/20 km/s

The derivative of the velocity is acceleration, The derivative of acceleration is jerk. So if we are asked to find the velocity and are given the acceleration all we need to do is integrate the acceleration.

Acceleration = -80000/x^2 integrate both sides
Velocity = -80000 * Integral(1/x^2)
= 80000/x + C.

We know that when x=6400 the velocity is 5 so we can find C from this.

5-80000/6400=C
=-7.5

So the velocity is V(x)=80000/x - 7.5

So when the rocket is 3600 above the radius x is then 3600+6400 = 10000

So V(10000)= 80000/10000 -7.5
=8-7.5
=.5 Km/s

----edit-----


My answer is correct
The other answer is off, its 80000 not 8000