Gravitation: Rocket's Velocity

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Homework Help Overview

The discussion revolves around a physics problem involving a rocket's motion in the context of gravitational forces. The original poster presents a scenario where a rocket, after reaching a certain altitude and velocity, has its engine shut off, prompting questions about its subsequent velocity and maximum height above the Earth's surface.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore various equations related to gravitational forces and energy conservation, questioning the validity of using escape velocity in this context. There are attempts to calculate the rocket's velocity at a higher altitude and its maximum height, with some participants expressing confusion over their calculations and the assumptions made.

Discussion Status

Several participants are actively engaging with the problem, providing different approaches and questioning each other's reasoning. There is no explicit consensus yet, but some guidance has been offered regarding the use of energy conservation principles and the need to consider gravitational potential energy.

Contextual Notes

Participants note the importance of understanding the changing gravitational force with altitude and the implications of using initial conditions for calculations. There is also mention of potential errors in calculations and assumptions that may affect the outcomes.

Juntao
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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)?

*********
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

I get answer of 108,240,434m/s

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. :(
 
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The first section is a little worrying first off. I would use the equation

[tex]F= \frac {GM_e}{R^2}[/tex]

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

[tex]v^2=u^2+2as[/tex]

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.
 
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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?
 
1.) You must remember that the acceleration is negative as the force is attracting the rocket to Earth, and [tex]v_0[/tex] 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 [tex]\frac{1}{R^2}[/tex] 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:
Originally posted by 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)?

*********
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

I get answer of 108,240,434m/s

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.

[tex] <br /> \Delta KE = \int F dr[/tex]

[tex] \Delta KE = \int_{200km}^{1,000km}-\frac{\mu}{r^2}dr<br /> [/tex]

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


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. :(

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

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