06 Apr 05
Originally posted by AcolyteHow much air does he have left?
Here's another bullet-themed puzzle:
An astronaut stranded on a desolate, airless planet plans to kill himself with a gun (which we'll assume is one which works in space) but can't bear to point the gun at himself. Instead, he carefully takes aim and fires the gun directly forward. The bullet travels at x m/s, and y seconds later, the bullet hits him i ...[text shortened]... uming the planet is spherical and of uniform density, what are the planet's radius and density?
06 Apr 05
Originally posted by AcolyteVery interesting. I had to brush up on some old physics to try answering it, which is always a good thing!
Here's another bullet-themed puzzle:
An astronaut stranded on a desolate, airless planet plans to kill himself with a gun (which we'll assume is one which works in space) but can't bear to point the gun at himself. Instead, he carefully takes aim and fires the gun directly forward. The bullet travels at x m/s, and y seconds later, the bullet hits him ...[text shortened]... uming the planet is spherical and of uniform density, what are the planet's radius and density?
1. Circumference/volume of the planet:
If the gun fires the bullet at x m/s, and y seconds later it makes a complete revolution of the planet, then it has travelled x*y metres. The circumference of the great circle it just travelled is 2*pi*r, so by equating those two expressions we get: r = x*y/(2*pi)
The volume of a sphere is 4/3*pi*r^3, so the volume of the planet is: 4/3*pi*[xy/(2pi)]^3.
2. Mass of the planet:
We know that the bullet is orbiting the planet at the height at which it was fired. For simplification purposes, let's assume this height is negligible compare to the radius of the planet. Centripetal acceleration provided by the gravitational force of the planet prevents the bullet from flying straight off into space, but it isn't so much that the bullet gets dragged into the surface of the planet.
centripetal force: Fc = m2*v^2/r
gravitational force: Fg = G*m1*m2/r^2
Equating these two we get: m2 = v^2*r/G
(where m2 is the mass of the planet)
In this problem, the velocity is x m/s, so we sub that in to get m2 = x^2*r/G.
3. Density of the planet
To find the density, we divide the mass from result 2 by the volume from result 1. Doing this and simplifying, we get:
rho = x^2/[4/3*G*pi*(xy/(2pi))^2]
It's always good to plug in some numbers to see what this looks like in real life. Let's assume the following values:
x = 1000 fps for a handgun (found on the internet) = 305 m/s
r = 6371 km (radius of the Earth) = 6371000 m
y = 131247 s (calculated) = 36.5 hours
G = 6.67E-11 Nm2/kg2
m2 = 8.81E+21 kg (calculated) = 0.15% of the Earth's mass
volume = 1.08E+21 m3
density = 0.0081 g/cm3 (calculated), which is about 6 times as dense as air at standard temperature and pressure
Interesting, no?
18 Apr 05
Originally posted by The PlumberNP, plumber. I often feel that I may understand, but don't fully comprehend solutions to physical problems unless I compare the numbers to something I know. I had an engineering professor who was an expert at explaining answers in terms of everyday objects (that's about as big as a swimming pool, as fast as a car travelling on the highway, as hot as a barbeque on HIGH, etc...). It really helped me develop physical intuition with complex systems.
Actually, yes - thanks for doing the research and calculation on that one.
18 Apr 05
Originally posted by PBE6I know what you mean. That explains why I was good at math right up until I started into partial differential equations - I just had a hard time relating those to the real world....
NP, plumber. I often feel that I may understand, but don't fully comprehend solutions to physical problems unless I compare the numbers to something I know. I had an engineering professor who was an expert at explaining answers in terms of everyday objects (that's about as big as a swimming pool, as fast as a car travelling on the highway, as hot as a barbeque on HIGH, etc...). It really helped me develop physical intuition with complex systems.
19 Apr 05
Originally posted by Palynkahi...you are quite corect. it depends on the velocity of the bullet completely(!). depending on it's velocity it may never fall in the earth to, it may orbit the earth(ignoring the air), or go out of earth. So the question is only defined, when we are ignoring the presence of the earth, but considering two planes: one plane and one curved- with the some mass...
Despite having no physics background(I'm a lowly economist):
Isn't it impossible to fire a projectile exactly parallel to the surface of the earth?
Then, the bullet would always be slightly rising (if assumed that he shot perpendicularly to current vertical position) as long as the horizontal velocity is higher than the vertical one.
I would expect the fired bullet to fall slightly later if this is true.
19 Apr 05
Bullet A will hit the water first, as you said that you drop Bullet B onto the raft, not the water, therefore because gravity is still in existance in the middle of the Pacific Ocean, Bullet A will eventually begin to drop and will hit the water before Bullet B, unless of course, Bullet B is picked up off the raft and dropped into the water before Bullet A gets there first.