What is the maximum height reached by a projectile fired from Earth?

[kraigandrews]

Homework Statement

A projectile of mass m is fired from the surface of the Earth at an angle α from the vertical, where α = 0.34 radians. The initial speed v0 is equal to √{GM/Rearth}. Calculate the maximum height that the projectile will reach. Neglect air resistance and the Earth's rotation. Express the result as the ratio to the radius of the Earth.

Homework Equations

U_G=(GMm)/R

KE=.5mv₀²

The Attempt at a Solution

So i believe that the fact that it is fired at an angle is irrelevant because Earth's gravitational field is in all directions.

Then i just equated the energy equations and solved for R, however this did not yield the correct answer. THoughts?

 

[PeterO]

The Attempt at a Solution

So i believe that the fact that it is fired at an angle is irrelevant because Earth's gravitational field is in all directions.

Then i just equated the energy equations and solved for R, however this did not yield the correct answer. THoughts?

If you get an incorrect answer, it usually means you have made at least one incorrect assumption.

The first assumption you made, was that the angle was irrelevant - so let's start there.

If you fire a gun vertically, the bullet will go very high, then eventually land at your feet [provided you have stepped back or it may land on your head]

When a marksman at the olympics fires a gun at the target, the bullet hardly rises and falls at all [probably a few cm only]

Those were two extreme examples where the angle makes a clearly noticable difference.

The angle is important.

btw: Gravity does not operate in all directions either - it acts towards the centre of the earth, which we on the surface usually call DOWN

Peter

 

[gneill]

I would be curious to know what particular course this problem comes from. It has elements that make it seem suited to a course on astrodynamics.

The initial speed given is that which would be held by an object in circular orbit at radius Rearth. That is, if it had been fired horizontally it would proceed to orbit the Earth at the radius of the surface in a circular path. This is less than escape speed, so as it's launched at an angle to the (local) horizontal, the projectile will describe an elliptical trajectory (no doubt intersecting the Earth's surface at some spot remote from the launch point -- can you say "ballistic missile"? )

The launch angle is going to affect the eccentricity of the orbit, so it will have an impact on the apogee. It looks like it will reach a height above the surface that will fall just short of an Earth radius.

I can think of a few "Relevant equations" from astrodynamics that would address this problem quite handily, but the question is, would they be relevant to your course of study?

 

[Dick]

I think you can manage this with just conservation of energy and conservation of angular momentum. That's got to be relevant to any course that would even ask this question.

 

[gneill]

I think you can manage this with just conservation of energy and conservation of angular momentum. That's got to be relevant to any course that would even ask this question.

True. The problem is amenable to solution using those principles.

Hints:
- The speed of the projectile at any radial distance r will depend upon its total mechanical energy.
- At apogee (and perigee), the velocity is perpendicular to the position vector.

 

[kraigandrews]

ok thank you for the hints. by the way, for those interested the class is classical mechanics

 

[kraigandrews]

The angle is really throwing me off in going at this problem.

 

[gneill]

Can you determine the total mechanical energy and the angular momentum (about the Earth's center) from the initial conditions?

 

[kraigandrews]

thanks for all the help i solved it, here's how:
using \alpha/r=1+Cos(\phi)
since it wants the height in terms of the Earth's radius i just did 1+cos(.34) and got the ratio.