# Gravitation: Rocket's Velocity

1. Jan 13, 2004

### Juntao

A 140 kg rocket is moving radially outward from the earth at an altitude of 200 km above the surface with a velocity of 3.5 km/sec. At this point, its final stage engine shuts off.

a)Ignoring any minor air resistance, what is the rocket's velocity 1000km above the surface of the earth?
b) What is the maximum height of the rocket above the earth's surface (using the initial rocket mass of 140 kg)?

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Part a was easy. Using the escape velocity equation, I did v=sqrt(2G*M/R)

where G is the 6.67E-11 constant
M is mass of earth
and R is radius of earth plus 100,000m

b) This is where I get stuck.

Using same equation
V=3.5km/sec=3500m/sec
G=6.67E-11
M=5.98E24kg
I solve for R, but its not working. I thought I'd take the R subtract earth radius from that, but its not working either. :(

2. Jan 13, 2004

### Kurdt

Staff Emeritus
The first section is a little worrying first off. I would use the equation

$$F= \frac {GM_e}{R^2}$$

to obtain a force to put in the $$F=ma$$ equation which is just the 10N as expected. Then put this into the calssic motion equation as follows

$$v^2=u^2+2as$$

which turns out to be 3.48km/s

for question 2 just consider that the kinetic energy of the rocket is being turned into gravitational potential energy. Set the two equations equal and rearrange to find R.

Last edited: Jan 14, 2004
3. Jan 14, 2004

### Juntao

1) Can you go over how'd you get that 3.48 km/sec please? I figured out the force, so it was 9.8N or 10 N, and using F=ma, I divided the force by mass of the rocket, to get an acceleration of .07m/sec^2

Then using that V^2=vo^2+2ax

a=.07
x= 1000km
v0= 0 ?
so then final V I get 11.8km/sec, which doesn't match 3.48 km/sec.

2) Ok, so 1/2mv^2=(-GMeM)/r
where v=3500m/sec
M=mass of rocket

So I solved for R, and subtracted the radius of the earth from that answer to get 58,750,979m, but its not working out right. Is that right, or is it just a bug in the computer system?

4. Jan 14, 2004

### Kurdt

Staff Emeritus
1.) You must remember that the acceleration is negative as the force is attracting the rocket to Earth, and $$v_0$$ as you have used is 3500m/s. Also the distance x would only be 800 000m as the rocket shuts down 200km above the Earth. This relation is crude as the acceleration due to the gravity changes very slightly with $$\frac{1}{R^2}$$ but it is a good approximation.

2.) For part two you have to remember that there is a "zero point" of gravitational potential as everyone on the surface of the earth feels because we are not at its gravitational centre. If You work out the gravitational potential at the surface of the earth and then add the kinetic energy to that value you should obtain the correct answer.

I apologise for my lack of earlier clarity. I hope this helps :)

Last edited: Jan 14, 2004
5. Jan 14, 2004

### enigma

Staff Emeritus
This should have gotten your "red alert" button flashing. The rocket is going 3,500m/s at an altitude of 200,000m and it's going 30,000 times faster when it gets to 1,000,000m?

I'm not sure why you used the escape velocity equation. I can't see how it would be valid in this problem.

Just do conservation of energy. You won't even need to do any approximations.

$$\Delta KE = \int F dr$$

$$\Delta KE = \int_{200km}^{1,000km}-\frac{\mu}{r^2}dr$$

You know the initial Kinetic energy, solve for the final kinetic energy.

mu is the gravitational parameter or M*G or 398,600km^3/sec^2

Solve similarly, except solve for an upper variable instead of 1000km.

Last edited: Jan 14, 2004