Guest Posted April 8, 2009 Report Share Posted April 8, 2009 Find out the time it takes for two bodies of mass M1 and M2 separated by distance D in space to Collide.? Don't forget to note that gravitational forces becomes stronger and stronger as the bodies approach closer and closer. F=G.M1.M2/D^2 a=F/m(HighSchool Physics equations Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 Find out the time it takes for two bodies of mass M1 and M2 separated by distance D in space to Collide.? Don't forget to note that gravitational forces becomes stronger and stronger as the bodies approach closer and closer. F=G.M1.M2/D^2 a=F/m(HighSchool Physics equations as long as there isn't a force that propels them towards each other, then they will never collide Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 Right, are we assuming an initial vector of 0 with no other gravitations effecting it? Otherwise it could go into a sustained orbit. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 as long as there isn't a force that propels them towards each other, then they will never collide That force would be gravity... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 That force would be gravity... Unless I'm incorrect there is no gravity in space. Am I right? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 Unless I'm incorrect there is no gravity in space. Am I right? FTW! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 FTW! I'm sorry what does that mean? A little slow today. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 I'm sorry what does that mean? A little slow today. Never mind. Correct me if wrong 'For The Win'? Thank you google! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 Never mind. Correct me if wrong 'For The Win'? Thank you google! yes sir. kinda like "in your face," but geekier. i agreed with your comeback; i still think the answer stands. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 yes sir. kinda like "in your face," but geekier. i agreed with your comeback; i still think the answer stands. oh ok. Thanks! I have just never seen that before so. Well thanx for backing me up! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 (edited) gravity still exists in space. Any object with mass has gravity. As the Earth orbits the Sun, the gravity of the sun counters the centripetal force of the rotation, keeping us where we are. Although I am wondering if there is some information that is supposed to be inferred, or if it is just a straight forward physics problem. If the solution is a straight forward calculation it will still be in terms of M1, M2, and D. So, do you mean for us to find an exp​ression for t? The question said to find t... Edited April 8, 2009 by wafflechip Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 oh ok. Thanks! I have just never seen that before so. Well thanx for backing me up! i'm thinking that since it is posed as a "logical" physical problem, the math is unnecessary and the answer is easy. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 Unless I'm incorrect there is no gravity in space. Am I right? Yes, there is gravity in space. That's what keeps the earth in orbit around the sun and the moon in orbit around the earth. You're confusing it with sound (sound waves can't travel in space because they need a medium, such as air). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 gravity still exists in space. Any object with mass has gravity. As the Earth orbits the Sun, the gravity of the sun counters the centripetal force of the rotation, keeping us where we are. Although I am wondering if there is some information that is supposed to be inferred, or if it is just a straight forward physics problem. I agree that near objects in space there is gravity near objects. However it doesn't say you are near objects. Therefore, without objects there is no gravity. That's how I look at it anyway. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 (edited) I agree that near objects in space there is gravity near objects. However it doesn't say you are near objects. Therefore, without objects there is no gravity. That's how I look at it anyway. It doesn't say they are far objects either. D could be 1,000,000 km or 0.000001 cm. That's why I think an expression is required. Distance does not affect gravity. All objects with mass have gravity. The questions is how the force "felt" by an object changes with distance and how that affects the time. Edited April 8, 2009 by wafflechip Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 I agree that near objects in space there is gravity near objects. However it doesn't say you are near objects. Therefore, without objects there is no gravity. That's how I look at it anyway. Yeah; it is just in space. If the question wants to add something like "and these objects are within planet x's gravitational field" and then give the coefficient, that would change the question. but unless newton's first law is a lie, we're gonna need more info to solve with actual numbers. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 every object that has mass has an infinite gravitational field. This means no matter the value of D , in space, theoretically, with the force of gravity between the objects, they would collide at some point. I still don't know the expression for how long this would take though. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 Yeah; it is just in space. If the question wants to add something like "and these objects are within planet x's gravitational field" and then give the coefficient, that would change the question. but unless newton's first law is a lie, we're gonna need more info to solve with actual numbers. The point is that both objects are within each other's gravitational field. In fact, if you think about it, all objects in the universe are within each other's gravitational field, it's just that the ones really far away from each other feel virtually no gravitational impact. For this problem, I think we'll need to calculate how the gravitational force will change over time by determining its change over distance in between the masses and incorporate the change in acceleration with changing force. This sounds like it'll need more than high school physics though. It will probably need some kind of differential equation which will then be integrated over time and/or distance. Not positive yet though. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 It should be possible to solve this in terms of G, D and M even though the answer will be pretty messy without numbers to plug into it. And it doesn't matter what the masses or the distance between them is, there will be a gravitational attraction. As far as we know, every object in the universe exerts a pull on every other object, but unless they are fairly close together, the force is pretty negligible. t = ((D^6)/(G^2 * M1 * M2))^(1/4) The equations given were F=G(M1 * M2 / D^2) and F=mg G and g are not equivalent, G is the gravitational constant while g is a local graviational acceleration. So, g in terms of G should be g = (m * G)/D^2 Also, the time it takes something to fall in a local gravitational field is t = (2D/g)^(1/2) Using those to equations with F=mg gives F=4((D^4)/((t^4)G)). Substituting that into F=G((M1 * M2)/D^2) and solving for t results in my answer. I'm not sure if that is correct, it's been a long time since Physics. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 It should be possible to solve this in terms of G, D and M even though the answer will be pretty messy without numbers to plug into it. And it doesn't matter what the masses or the distance between them is, there will be a gravitational attraction. As far as we know, every object in the universe exerts a pull on every other object, but unless they are fairly close together, the force is pretty negligible. t = ((D^6)/(G^2 * M1 * M2))^(1/4) The equations given were F=G(M1 * M2 / D^2) and F=mg G and g are not equivalent, G is the gravitational constant while g is a local graviational acceleration. So, g in terms of G should be g = (m * G)/D^2 Also, the time it takes something to fall in a local gravitational field is t = (2D/g)^(1/2) Using those to equations with F=mg gives F=4((D^4)/((t^4)G)). Substituting that into F=G((M1 * M2)/D^2) and solving for t results in my answer. I'm not sure if that is correct, it's been a long time since Physics. This solution may be correct, but one thing that doesn't fit with me is ...g is a local graviational acceleration. So, g in terms of G should be g = (m * G)/D^2 I'm assuming you mean local gravitational acceleration similar to that of Earth, where g=9.8 m/s2. However, that wouldn't be the situation for two unknown masses. The gravitational force the Earth imposes actually isn't constant; it increases the closer we are to the center. However, since the radius of the earth is so large in comparison to even the diameter of earth's atmosphere, we consider g a constant. Thus, g will not not be constant because D is constantly changing. I, too, seem to be having difficulty solving what one might think would be a straightforward problem. I think I have some equations in place, but they involve some advanced calculus which I've forgotten since my college years. D(t) = D - d1(t) - d2(t) ---Where D(t) is the distance between the two objects at time t, d1(t) and d2(t) is the distance the first and second objects have traveled at time t. a1(t) = (m2*G)/D(t)2 ---Where a1 is the acceleration of the first object at time t. a2(t) can be found in the same manner. v1(t) = 0∫ta1(t) dt ---Where v1 is the velocity of the first object at time t. v2(t) can be found in the same manner. d1(t) = 0∫tv1(t) dt Thus d1(t) = (m2*G)0∫t(0∫t1/(D - d1(t) - d2(t))2 dt) dt Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 8, 2009 Report Share Posted April 8, 2009 This solution may be correct, but one thing that doesn't fit with me is ...g is a local graviational acceleration. So, g in terms of G should be g = (m * G)/D^2 I'm assuming you mean local gravitational acceleration similar to that of Earth, where g=9.8 m/s2. However, that wouldn't be the situation for two unknown masses. The gravitational force the Earth imposes actually isn't constant; it increases the closer we are to the center. However, since the radius of the earth is so large in comparison to even the diameter of earth's atmosphere, we consider g a constant. Thus, g will not not be constant because D is constantly changing. I mean local gravitational accelaration for any body, including the two in the problem. But you'll notice that the equation takes into account both the mass of the body and the distance. It's correct. That's not the part that I was concerned about when I said it might be wrong. I'm not sure if I'm allowed to substitute in such a way that I eliminate m and then go back and use an equation containing M1 and M2. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 9, 2009 Report Share Posted April 9, 2009 To be honest, I don't know the anwer. I thought since the question is simply stated, the answer might be simple. And I have realized that its not As simple as determinig the time for when objects falling on the earth hits the ground. There is basically two difficulties this time----> Both the masses accelerate and the acceleration changes with distance. So I souppose It must involve dificult calculus. I am not trying to make you do the calculus, but since I myself thought the solution could be simple, just posted it if anyone of you could solve it. What Vinays84 has done is the answer, leaving full math. But there is diffulty in Integrating acclereation with respect to time since we have its expression that varies with distance. However we aren't concerned with technical details about solving this difficulty, I take your answer as correct Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 9, 2009 Report Share Posted April 9, 2009 There is basically two difficulties this time----> Both the masses accelerate and the acceleration changes with distance. The former is not problematic, because we only care about their relative distance. The real problem is that the DE is non-linear: r'' = -G(m1+m2)/r2 That's not a form I recognize, which doesn't mean it doesn't have a solution. However, DE's were never exactly my forte, so I'll let the hard-core math geeks handle it from here. The other avenue is to consider: 1/2 (m1+m2)(dr/dt)2 - GMm/r = -GMm/r0 but at first glance, that doesn't appear to help. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 10, 2009 Report Share Posted April 10, 2009 The former is not problematic, because we only care about their relative distance. The real problem is that the DE is non-linear: r'' = -G(m1+m2)/r2 That's not a form I recognize, which doesn't mean it doesn't have a solution. However, DE's were never exactly my forte, so I'll let the hard-core math geeks handle it from here. The other avenue is to consider: 1/2 (m1+m2)(dr/dt)2 - GMm/r = -GMm/r0 but at first glance, that doesn't appear to help. You are almost there. Starting with your differential equation: r'' = -G(m1+m2)/r2 Multiply by r2 to get: r2(d2r/dt2) = -G(m1+m2) Multiply by dt2: r2(d2r)= -G(m1+m2)dt2 Integrate the equation twice: r4/12 = -G(m1+m2) t2/2 Evaluate the integral on the interval [(0,tf),(D,0)] D4/6 - 0 = 0 + G(m1+m2) tf2 Solve for tf: tf = (D2)/(6G(m1+m2))1/2 I hope I did the math correctly. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 10, 2009 Report Share Posted April 10, 2009 You are almost there. Starting with your differential equation: r'' = -G(m1+m2)/r2 Multiply by r2 to get: r2(d2r/dt2) = -G(m1+m2) Multiply by dt2: r2(d2r)= -G(m1+m2)dt2 Integrate the equation twice: r4/12 = -G(m1+m2) t2/2 Evaluate the integral on the interval [(0,tf),(D,0)] D4/6 - 0 = 0 + G(m1+m2) tf2 Solve for tf: tf = (D2)/(6G(m1+m2))1/2 I hope I did the math correctly. Looks good to me. I somehow failed to notice it was separable... Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 12, 2009 Report Share Posted April 12, 2009 Gravity's force is inverse square law of the separation of the centers of mass. But objects don't collide when their centers of mass coincide, they collide when their external surfaces come into contact. For the problem to be well posed, we need something like: The centers of two spheres, of mass m1 and m2 and radius r1 and r2, initially at rest, are separated by a distance d. Under the force of their gravitational attraction, at what time will they collide? The answer then is not when d becomes zero but when it becomes r1 + r2. Quote Link to comment Share on other sites More sharing options...
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Find out the time it takes for two bodies of mass M1 and M2 separated by distance D in space to Collide.? Don't forget to note that gravitational forces becomes stronger and stronger as the bodies approach closer and closer.
(HighSchool Physics equations
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