Some fun Physics questions

[Cogman]

First couple come from me watching Space cowboys . Ok, so the chick said that as long as you hit a baseball halfway to the moon, lunar gravity will take over and bring it in all the way. My instant reaction is "Thats not true!" Lunar gravity is much less then earths gravity therefore the distance must be greater then 1/2 for the moon to take over. So here is the challenge, calculate how far the baseball has to go (assuming standard distance from the moon to the earth) before the moons gravity will take over (so calculating where unstable equilibrium is). And if you really want, the force that must be applied on a .1 kg baseball to get it there.

Next, she said that if you bounced on a trampoline that you would just go up forever, Of course I said "Thats not true!" So assume there is nothing in the universe except for you and the trampoline, you have a mass of 100 kg an the trampoline a mass of 150 kg. If you exerted a force of 500 N (over 1 second) on the trampoline, how far would you go until you stopped moving and long would it take for you and the trampoline to make contact once again?

Next, not a math problem, Our Physics teacher was teaching us about the hall effect, he told us that it could be used to monitor blood flow and see if there is a restriction anywhere. Well, he also told us we need a volt meter probe connected to the two sides of the arteries. Being the squeamish type that I am, I don't like the idea of having needles jabbed multiple times just to see if my blood is flowing (though I could see the use in open heart surgery). So is there another way to delicately measure a voltage across a plane? or in other works, another way to use the hall effect to measure blood flow in arteries? Just curious.

Umm, one more, How exactly does sending a current around in a circular path case a magnetic field?

Thanks, and good luck to those that wish to answer

 

[foges]

first question: i think its around 7/8 of the distance to the moon.

Edit: Upon calculation it seems that it must be ~9/10 of the distance from the earth's center to the moon's center

second question: im prety sure that its true, well as long as you bounce hard enough, think about the escape velocity from earth or any planet/moon/mass.

third question: dont know what the hall effect is

fourth question: i think thats one of those wonders of the universe, just like why does any mass attract another, beats me, the part that interests, why does a current (conventional) create a magnetic field in the anti-clockwise direction, why not clockwise. (i guess its like asking why do masses attract eachother and not repulse eachother

 

[Mark R]

3. I can't see any other way in which the Hall effect would be useful.

In the classical hall effect electrical charge carriers are pushed through a conductor by an external current source. At right angles, a magnetic field acts causing deflection of the charge carriers. So, a voltage develops across the conductor in an orthogonal direction.

Presumably, for flow assessment, this would be done by replacing the current source, with physical movement of the blood.

There is an experimental technique for measuring the dielectric coefficient of tissues using the Hall effect, in this case the movement comes from ultrasound. However, this is hardly an established, nor even practical, technique.

There are better ways of measuring flows through arteries, specifically Doppler ultrasound. This not only demonstrates the anatomy of the artery, but also flow characteristics: flow velocity, resistance to flow and turbulence.

 

[Born2bwire]

Originally posted by: foges
first question: i think its around 7/8 of the distance to the moon.

Edit: Upon calculation it seems that it must be ~9/10 of the distance from the earth's center to the moon's center

second question: im prety sure that its true, well as long as you bounce hard enough, think about the escape velocity from earth or any planet/moon/mass.

third question: dont know what the hall effect is

fourth question: i think thats one of those wonders of the universe, just like why does any mass attract another, beats me, the part that interests, why does a current (conventional) create a magnetic field in the anti-clockwise direction, why not clockwise. (i guess its like asking why do masses attract eachother and not repulse eachother

Not necessarily. Let's say we accelerate away from the Earth at g. Then, halfway to the moon we stop accelerating. Now the acceleration towards the Earth is going to be distance dependent, but less than g. So assuming -g acceleration, we will arrive to a stop at the moon. It isn't a case of when the moon's force is greater, but one of work. I'd write more but I am typing this on my Wii.

 

[firewolfsm]

For two, I think gravity would eventually bring you and trampoline back to each other, it would take a long ass time though.

 

[earthman]

The second scenario doesn't make any sense. If you were in space in an essentially zero-g situation, you couldn't "bounce" on a trampoline, if you tried to jump off it you would keep traveling away and there would be no "bounce". The two objects would move apart at differing speeds depending on their initial mass.

In the first scenario, if you hit a baseball hard enough to escape earth's gravity, the gravity of the moon would be irrelevant, as the baseball would be traveling much faster than any acceleration it would ever get from the moon. The moon is only .0123 the mass of earth, or something similar. The Lagrange point between the earth and moon is about 85% of the way there.

 

[Mark R]

1. I get 90% of distance between earth and moon centres.

Distance between E & M = 1 unit
Distance from E to object = a
Mass of moon = kM = 0.0123 M
Force exerted by earth = GM/a^2
Force exerted by moon = -kGM/(1-a)^2

Rearrange and simplify: 1 - 2a + (1-k) a^2 = 0
Solve: a = 0.900 or 1.112 (latter is meaningless)

As the moon's gravitation energy (mass) is so much smaller than the earth's (about 1%), it makes very little difference to the energy required to escape from the earth's gravity.

Earth escape velocity is 11.2 km s-1; to get to the moon the escape velocity is 11.1 km s-1.

 

[Born2bwire]

Originally posted by: Mark R
1. I get 90% of distance between earth and moon centres.

Distance between E & M = 1 unit
Distance from E to object = a
Mass of moon = kM = 0.0123 M
Force exerted by earth = GM/a^2
Force exerted by moon = -kGM/(1-a)^2

Rearrange and simplify: 1 - 2a + (1-k) a^2 = 0
Solve: a = 0.900 or 1.112 (latter is meaningless)

As the moon's gravitation energy (mass) is so much smaller than the earth's (about 1%), it makes very little difference to the energy required to escape from the earth's gravity.

Earth escape velocity is 11.2 km s-1; to get to the moon the escape velocity is 11.1 km s-1.

Again, you're forgetting that you need to slow the craft down and then reverse its velocity before you can say that the craft won't make it to the moon.

Sec, let me run some numbers.

EDIT: My quick calculations (which may be prone to math errors) suggest that if you can keep your craft at about 3.6 km/s for half the distance from the Earth to the moon you should make it. Of course, this doesn't work since you need 11.2 km/s to escape the Earth's velocity. Basically, I calculated that the work that must be done against the Earth and moon is 1.2404e6*mass (in terms of N-m). So you would need to expend at least this amount of energy to move from the Earth to the moon. My calculation does not work because the one caveat is that you need to exert enough force to overcome the force of the Earth minus the force of the Moon. So as long as your applied force is greater than the force back to the Earth, you should only need to apply this force for 1.2404e6*mass/F = 1.2404e6/a meters. So if you exerted an acceleration of g, then you would only need to apply the force for around 126,600 km, less than halfway.

 

[Born2bwire]

Originally posted by: Cogman
Umm, one more, How exactly does sending a current around in a circular path case a magnetic field?

You don't need to force the current on a circular path to produce a magnetic field. A straight wire of constant current will produce a magnetic field as a consequence of Maxwell's Equations, more specifically via Ampere's Law.

 

[CTho9305]

EDIT: My quick calculations (which may be prone to math errors) suggest that if you can keep your craft at about 3.6 km/s for half the distance from the Earth to the moon you should make it. Of course, this doesn't work since you need 11.2 km/s to escape the Earth's velocity.

No you don't. You only need to go 11.2km/s if you're starting at the Earth's surface and don't have any further forces applied (i.e. you turn off the rocket engine). If you could build a tall enough elevator/ladder/whatever, you could take your time going to the moon.

 

[earthman]

I think most of these scenarios ignore the fact that paths between planets are orbits, not straight lines. The moon is moving about 1 mile/sec in its orbit, so which direction you approached it from could make a difference.

 

[silverpig]

Originally posted by: Born2bwire

Originally posted by: Mark R
1. I get 90% of distance between earth and moon centres.

Distance between E & M = 1 unit
Distance from E to object = a
Mass of moon = kM = 0.0123 M
Force exerted by earth = GM/a^2
Force exerted by moon = -kGM/(1-a)^2

Rearrange and simplify: 1 - 2a + (1-k) a^2 = 0
Solve: a = 0.900 or 1.112 (latter is meaningless)

As the moon's gravitation energy (mass) is so much smaller than the earth's (about 1%), it makes very little difference to the energy required to escape from the earth's gravity.

Earth escape velocity is 11.2 km s-1; to get to the moon the escape velocity is 11.1 km s-1.

Again, you're forgetting that you need to slow the craft down and then reverse its velocity before you can say that the craft won't make it to the moon.

Sec, let me run some numbers.

EDIT: My quick calculations (which may be prone to math errors) suggest that if you can keep your craft at about 3.6 km/s for half the distance from the Earth to the moon you should make it. Of course, this doesn't work since you need 11.2 km/s to escape the Earth's velocity. Basically, I calculated that the work that must be done against the Earth and moon is 1.2404e6*mass (in terms of N-m). So you would need to expend at least this amount of energy to move from the Earth to the moon. My calculation does not work because the one caveat is that you need to exert enough force to overcome the force of the Earth minus the force of the Moon. So as long as your applied force is greater than the force back to the Earth, you should only need to apply this force for 1.2404e6*mass/F = 1.2404e6/a kilometers. So if you exerted an acceleration of g, then you would only need to apply the force for around 126,600 km, less than halfway.

You're not doing it right. You don't accelerate along the way, you don't fire an engine to maintain velocity along the way. You just hit the ball at some velocity v, let the earth and moon pull on it and see what happens.

 

[lousydood]

For #1 I think what you want is the Earth-Moon LaGrange point L1. Wikipedia says it's about 56000km from the surface of the Moon, which would make it about 325000km from Earth. An object in L1, assuming a 3-body problem, would be balanced by all forces and remain at rest relative to the Earth-Moon system. In reality, there are other bodies and effects in the Solar system, so craft at LaGrange points need to maintain some kind of esoteric orbit around the point itself.

 

[Mark R]

Originally posted by: lousydood
For #1 I think what you want is the Earth-Moon LaGrange point L1. Wikipedia says it's about 56000km from the surface of the Moon, which would make it about 325000km from Earth. An object in L1, assuming a 3-body problem, would be balanced by all forces and remain at rest relative to the Earth-Moon system. In reality, there are other bodies and effects in the Solar system, so craft at LaGrange points need to maintain some kind of esoteric orbit around the point itself.

No. The problem is asking for the point at which the net gravitational field between the two bodies is zero.

This is not the L1 point. At L1 the gravitational field is such that orbital time would be the same for this narrower orbit as for the moon.

 

[KIAman]

Originally posted by: firewolfsm
For two, I think gravity would eventually bring you and trampoline back to each other, it would take a long ass time though.

Gravity is pretty weak and its force is inversely proportional to the square of the distance between 2 bodies.

So we got 2 F(gravity) = G(m1*m2)/r^2 (G ~ 6.67428 * 10^-11 Nm^2kg^-2)

So F = G(100*150)/r^2

At close to contact (1mm), the 150kg body and 100kg trampoline only exerts 1N of gravitation force on each other. Once the distance reaches a full meter the force of gravity is 1x10^-6N.

Assuming a 500N force at contact, the acceleration would easily push the trampoline and the body past the point where gravity could ever bring them together.

The only way to do that is to have an attraction at 500N of force would be to have the 150kg trampoline with a 50,000kg person.

Consider why the universe is pulling apart and to support a repeating universe lifecycle, we had to add a bunch of matter to make the equation work. Considering only the visible matter, the universe will blow itself apart forever.

 

[Nathelion]

Originally posted by: KIAman

Originally posted by: firewolfsm
For two, I think gravity would eventually bring you and trampoline back to each other, it would take a long ass time though.

Gravity is pretty weak and its force is inversely proportional to the square of the distance between 2 bodies.

So we got 2 F(gravity) = G(m1*m2)/r^2 (G ~ 6.67428 * 10^-11 Nm^2kg^-2)

So F = G(100*150)/r^2

At close to contact (1mm), the 150kg body and 100kg trampoline only exerts 1N of gravitation force on each other. Once the distance reaches a full meter the force of gravity is 1x10^-6N.

Assuming a 500N force at contact, the acceleration would easily push the trampoline and the body past the point where gravity could ever bring them together.

The only way to do that is to have an attraction at 500N of force would be to have the 150kg trampoline with a 50,000kg person.

Consider why the universe is pulling apart and to support a repeating universe lifecycle, we had to add a bunch of matter to make the equation work. Considering only the visible matter, the universe will blow itself apart forever.

The universe IS blowing itself apart forever (or at least that's the current theory), and this observation is what has prompted the postulation of dark energy. Not the other way around. There is no law saying that the universe must have a recurring "life cycle".

 

[silverpig]

Originally posted by: KIAman
Consider why the universe is pulling apart and to support a repeating universe lifecycle, we had to add a bunch of matter to make the equation work. Considering only the visible matter, the universe will blow itself apart forever.

Actually dark matter was added to solve the rotation curve problem of galaxies. IE: the outer stars of galaxies don't act like they're on the edge of something massive, they act like they're in the MIDDLE of something massive. That is, there must be a lot more mass outside the outermost stars that we can't see.

Dark energy is added for a number of reasons, mainly to explain the accelerating expansion of the universe.

 

[DrPizza]

trampoline question above - you could easily reach escape velocity.

 

[Born2bwire]

Originally posted by: silverpig

Originally posted by: Born2bwire

Originally posted by: Mark R
1. I get 90% of distance between earth and moon centres.

Distance between E & M = 1 unit
Distance from E to object = a
Mass of moon = kM = 0.0123 M
Force exerted by earth = GM/a^2
Force exerted by moon = -kGM/(1-a)^2

Rearrange and simplify: 1 - 2a + (1-k) a^2 = 0
Solve: a = 0.900 or 1.112 (latter is meaningless)

As the moon's gravitation energy (mass) is so much smaller than the earth's (about 1%), it makes very little difference to the energy required to escape from the earth's gravity.

Earth escape velocity is 11.2 km s-1; to get to the moon the escape velocity is 11.1 km s-1.

Again, you're forgetting that you need to slow the craft down and then reverse its velocity before you can say that the craft won't make it to the moon.

Sec, let me run some numbers.

EDIT: My quick calculations (which may be prone to math errors) suggest that if you can keep your craft at about 3.6 km/s for half the distance from the Earth to the moon you should make it. Of course, this doesn't work since you need 11.2 km/s to escape the Earth's velocity. Basically, I calculated that the work that must be done against the Earth and moon is 1.2404e6*mass (in terms of N-m). So you would need to expend at least this amount of energy to move from the Earth to the moon. My calculation does not work because the one caveat is that you need to exert enough force to overcome the force of the Earth minus the force of the Moon. So as long as your applied force is greater than the force back to the Earth, you should only need to apply this force for 1.2404e6*mass/F = 1.2404e6/a kilometers. So if you exerted an acceleration of g, then you would only need to apply the force for around 126,600 km, less than halfway.

You're not doing it right. You don't accelerate along the way, you don't fire an engine to maintain velocity along the way. You just hit the ball at some velocity v, let the earth and moon pull on it and see what happens.

Right, I see what you guys mean. I misread it, I thought that the statement was that they apply a series of burns or some periodic or effective force until they reach the halfway point, at which they do not apply anymore forces. Taking the statement that they only apply enough force to get them to the halfway point where the velocity would reach zero is wrong and should be 9/10 of the way instead.