Projectile Range Equation.

[Gareth56]

Here’s part of a problem from Tipler Physics for Scientists & Engineers:-

 

In a Sci-fi short story written in the 70s, Ben Bova described a conflict between two hypothetical colonies on the Moon- one founded by the USA the other by the USSR. In the story, colonists are from each side started firing bullets at each other, only to find to their horror that their rifles had a high enough muzzle velocity that the bullets went into orbit. (a) If the magnitude of free-fall acceleration on the Moon is 1.67m/s^2 , what is the maximum range of a rifle bullet with a muzzle velocity of 900m/s? (assume the curvature at the surface of the Moon is negligible).

 

 

I read this as being what is the horizontal range thus using the Range formula:

 

R = U^2 sin(2alpha)/g where g = 1.67m/s^2 U = 900m/s

 

I used zero degrees in my answer for the reason that the rifle was held horizontally with respect to the surface of the Moon. However in the answer the angle in the Range formula was 90degrees. Surely that’s pointing the rifle straight up with respect to the surface of the Moon.

 

Have I misunderstood the question? If so why is the angle 90degrees?

[Sisyphus]

What does the rest of the problem say?

[Gareth56]

Part (b) What would the muzzle velocity have to be to send the bullet into circular orbit just above the surface of the Moon? (You will need to look up the radius of the Moon)

 

For interest it's Q95 of Chapter 3 of the book.

[Sisyphus]

I don't understand why it would be either horizontal or vertical. At horizontal, the range would be zero. And at vertical, it's just coming back down on your head. But since they bother to mention curvature, they can't be looking for vertical height. I would have to assume they'd be talking about a 45 degree angle, which will give you the maximum horizontal range.

[Gareth56]

The answer part (a) according to the book is 485km. They have used 90deg as the angle in the range equation which to me is pointing straight up into space (if your standing on the Moon).

 

What I don't understand is why haven't they taken the horizontal range which is the normal way (I think) when the angle between the flight of the bullet and the ground would be zero.

[Sisyphus]

What I don't understand is why haven't they taken the horizontal range which is the normal way (I think) when the angle between the flight of the bullet and the ground would be zero.

 

Wouldn't that make the range also zero? They don't give any initial height that it's fired from, right? So you have to assume it's the surface. So if there's no vertical component, the instant it leaves the rifle it hits the ground, because it's falling at 1.67m/s^2 from a height of zero. Range: zero. Am I missing something? What was your answer, and how did you arrive at it?

[swansont]

90º in the calculation, but note that the formula uses [math]2\theta[/math], so the elevation angle is 45º, which is exactly what you'd expect for maximum range.

[Gareth56]

Oh no how embarrassing. I'll go and stand in the corner

[swansont]

No need to feel embarrassed. Spotting mistakes is often a function of having seen or made them, i.e. much more a function of experience than intelligence. And you shouldn't feel a need to apologize (IMO) for not having a great deal of experience.

[blazarwolf]

^

l

l

Exactly!, I thought for a good 10 seconds "but what kinda bullet" before I realized the moon has very little atmosphere. Then thought "but how did combustion take place" before I realized cartriges are sealed... then I returned to an idea I think ive had before... compressed pure oxygen in rifle cartriges... ofcourse this is just a painfull method of amputation... so dont worry about it, unless that corner is your special thinking place .