Motion of a rocket after all the fuel is exhausted?

[Eunjung]

Homework Statement

Hello, I am currently a gr 12 high school student, and I am stuck in a problem. Please help :)

The problem: Consider the motion of a rocket near the surface of the Earth. In this case, mass, in the form of gas is expelled from the rocket to propel the rocket upwards. A simple model of this would be the ejection of cannonballs from the back end of a frictionless cart initially at rest. What happens to the rocket when all the fuel is exhausted?

i) Explain the physics of the problem
ii) Graphical output to display variables that are relevant to the problem.

Homework Equations

Force of gravity (global) = G * Mass of Earth * Mass of rocket / (Earth's radius + distance rocket travels above the surface)^2

G= 6.67*10^-11 Nm^2/kg^2
Mass of Earth = 5.98 * 10^24 kg
Mass of rocket = unknown
Earth's radius = 6.4 x 10^6 m

The local gravitational force formula F = mg

The Attempt at a Solution

I got up to the point where the rocket goes up for a bit, due to inertia, which may carry it outside of the Earth's atmosphere. But, what if the rocket can't escape the atmosphere?

The rocket will be at 0 velocity at the peak of its motion, but I cannot think of how to calculate the acceleration in the rocket's way down.

The gravitational force grows weaker as you get further away from the surface of the Earth, so it can't possibly be F = ma (the force of gravity on the rocket will constantly change, so I don't think I can use that one since it generalizes everything).

I tried using the global gravitational force formula, but I don't know the maximum height of the rocket, so I can't calculate the force of gravity at different heights. Will it not make a difference because of the large radius of the Earth? I don't know.

If I assume that the mass of the rocket and the height in which the rocket flies above the surface is insignificant, then the force of gravity will be 9.7379N for the entire downward motion of the rocket. Gravity changes as you go out of Earth's atmosphere, so that can't be right.

I'm totally lost, can someone please help?

-Eunjung :)

 

[NEILS BOHR]

after all the fuel is xhausted , the motion of the rocket wud be FREE FALL..

 

[Eunjung]

Will it be regular free fall?

I think free fall assumes the same gravitational force for the entire downward motion.

However, rockets go high enough to feel a different gravitational force at the top of its motion. As it goes down, it will experience more and more gravitational force.

So I don't think it will be a normal free fall motion.

 

[tiny-tim]

Welcome to PF!

Hi Eunjung! Welcome to PF!

… So I don't think it will be a normal free fall motion.

It will be in orbit.

The orbit will have a very high eccentricity (very elongated), and will intersect the Earth's surface.

 

[Eunjung]

Sorry, but can you briefly explain to me why it would be an elongated orbit?

I don't quite understand how orbits work. Sorry.

 

[tiny-tim]

An orbit just means an ellipse (including a circle).

Any ellipse can be an orbit.

Sir Isaac Newton worked out that if he threw an apple fast enough horizontally, then (ignoring air resistance and things like trees in the way) the apple would orbit the Earth continuously in a circle (assuming he got out of the way!).

Of course, if he threw it at a slight angle upward at the same speed, it would still go most of the way round the Earth, but it would crash just before it got back.

However, if he stood on top of a tall tower, he wouldn't have to throw it exactly horizontally …

and the taller the tower, the greater the angle at which he could launch it into orbit (if he threw it fast enough).

Whatever the initial velocity of a projectile (less than escape velocity ), it will follow an ellipse …

the only question is whether that ellipse intersects the Earth's surface.

 

[Eunjung]

But since the rocket is being shot upwards, probably at 90 degrees, it will probably hit the surface of the Earth. The angle is high enough I think.

But, assuming that the rocket does hit the surface, how would I calculate the acceleration or velocity at a certain height of the rocket on its way down? I am not given the mass of the rocket or anything...

By the way, thanks a lot for your help :)

 

[ehild]

Eunjung, the problem can not be so complicated if you are in the 12th class. Read the text of the problem carefully:

"consider the motion of the rocket near the surface of the Earth"

It means the rocket does not go out from the atmosphere, does not go to orbit around the Earth, it can be some simple rocket used to signal, to warn, or to play at New Year's Eve. You can assume constant gravitational acceleration.

"mass, in the form of gas is expelled from the rocket to propel the rocket upwards."

So the rocket moves vertically upwards.

The rocket gains speed to move upwards from the ejected gas, according to Newton's third law/ conservation of momentum.

When all the fuel exhausted, only the bare shell remains moving upwards, like a stone. What happens with a stone when thrown upwards? What is its acceleration, how does the velocity and height change with time?

ehild

 

[tiny-tim]

Will it be regular free fall?

I think free fall assumes the same gravitational force for the entire downward motion.

However, rockets go high enough to feel a different gravitational force at the top of its motion. As it goes down, it will experience more and more gravitational force.

So I don't think it will be a normal free fall motion.

It will be in orbit.

The orbit will have a very high eccentricity (very elongated), and will intersect the Earth's surface.

But since the rocket is being shot upwards, probably at 90 degrees, it will probably hit the surface of the Earth. The angle is high enough I think.

But, assuming that the rocket does hit the surface, how would I calculate the acceleration or velocity at a certain height of the rocket on its way down? I am not given the mass of the rocket or anything...

As ehild says, the question seems to intend that you assume a constant acceleration of -g.

However, since you raise the topic of experiencing more and more gravitational force, you can still solve it using conservation of energy …

instead of using potential energy = mgh, you use -mgR/r, where R is the Earth's radius, and r is the height of the rocket above the centre of the Earth (and the acceleration instead of -g is -gR²/r²).

You don't need to know what m is, because it cancels (with the m in 1/2 mv², the kinetic energy ).

(and yes, if it's almost 90° then it's bound to crash!)

 

[Eunjung]

Oh, it's that simple?

Haha, I was thinking rocket science, so it's bound to be complicated. But I guess not..

So it's just simple conservation of mechanical energy? Well, except I need to use the -mgR/r as potential energy..

Thank you!