Jump to content

Calculator S.O.S.


fancycat

Recommended Posts

So I am working on some college physics homework but my calculator is giving me an answer that is off by over 10 million.

my equation is sq root (2GM(1/a -1/b)) where G = 6.67*10^-11 M= 2.5*10^30 a= 200000 and b = 200000+2.5*10^10

the confirmed correct answer is 28'874'613.94 however my calculator gives me an answer that is over 40 million. can anyone please help me figure out what is wrong.

I have also plugged my formula into wolfram alpha and got over 40 million. a friend with a graphing calculator got the correct answer but I am not allowed to use one of those. I have tried braking it up and doing it one piece at a time and i have tried doing it as one chunk bot times I got over 40 million. a friend with a standard calculator also gets over 40 million.

I have been working on this for 6 hrs on friday and 3 hrs today. if anyone can help me figure out how to do this on a standard calculator i would be eternally grateful.

my calculator: Sharp EL-531X

Link to comment
Share on other sites

yes it is a problem dealing with the velocity of two spheres that start at rest at a distance 2.5*10^10 the radius of the spheres are both 100000 each so starting distance is 2xrxd where d is the distance from surface to surface. because it is a gravity problem i need to measure distance between the centers of each sphere.

Link to comment
Share on other sites

Wherever you got that 28'874'613.94 number, it is not the correct answer for the formula you provided. Even doing this by hand you can see that the answer is about 4x10^7. The 1/b term is absolutely irrelevant, because it is much, much smaller than 1/a. So the whole thing becomes Sqrt(2 * G * M / a) = Sqrt(2 * 6.67x10^-11 * 2.5x10^30 / 2x10^5) = Sqrt(6.67*2.5x10^14) and that's roughly Sqrt(16x10^14) = 4x10^7.

Don't over-rely on calculators. Learn to do estimates without them. So either the "correct answer" is wrong, you have the wrong formula, or one of the parameters is wrong. I'd suspect a, because the formula is obviously for the orbit around a star comparable to Sun, and nothing's getting within 200,000m of its center. (If this is for orbit around black hole or neutron star, the formula is wrong to begin with.)

Edit: If this is for two objects orbiting a barycenter, have you taken reduced mass into account? It'll make a difference by root(2) in this case, which exactly accounts for the difference.

Edited by K^2
Link to comment
Share on other sites

this is a webassign homework problem that reads...

"Two neutron stars are separated by a distance of 2.5x10^10 m. They each have a mass of 2.5x10^30 kg and a radius of 1.0x10^5 m. They are initially at rest with respect to each other. As measured from that rest frame, how fast are they moving just before collision."

the answer that the system gave me as the right answer is 28'874'613.94 m/s this is the correct answer for the problem according to the answer key. the equation given by the answer key as the correct way to solve this is the one I provided above.

Link to comment
Share on other sites

yes it is a problem dealing with the velocity of two spheres that start at rest at a distance 2.5*10^10 the radius of the spheres are both 100000 each so starting distance is 2xrxd where d is the distance from surface to surface. because it is a gravity problem i need to measure distance between the centers of each sphere.

OK… so you have two sphere, each with a radius of 100,000… that's 100,000 what? Meters? Kilometers? Something else? If you are using G = 6.673e-11, then all your units had better be in kilograms, meters, and seconds (or combinations based on those). If they start at rest at a distance of d = 2.5e+10 meters apart… then that's a separation between their centers of 2 * r + d = 2 * 1e+5 meters + 2.5e+10 meters = 2.5e+10 meters, close enough. On that scale, the radius doesn't make any difference.

Now… if you release those from rest, they should "fall together", gaining velocity the whole way down, right? That sounds like a problem with escape velocity or energy conservation… you are applying an equation (often called the "Vis-Viva" equation in physics) that considers energy conservation but deals with the case where one mass is much much more massive than the other. Is that the case here?

You may or may not have yet another problem - how are you entering numbers into your calculator? It may not matter here, but scientific notation has a different order of operations than the "normal" functions like multiplication… so 1000 / 100 = 10, but 1*10^3/1*10^2 = 100,000, and (on my calculator), 1e+3 / 1e+2 = 10 again. Do you see the difference? Do you know which your calculator is doing?

Yeah, that Vis-viva equation is handy - it's how I figure out my transfer orbits in KSP all the time.

--

Brian Davis

Link to comment
Share on other sites

Yup. You forgot the reduced mass.

there is no change in mass. both objects have the same mass which is a constant.

....

with 1e^3 / 1e^2 = 10 on my calculator

this is a physics 1 class so some stuff isn't present. G=6.67x10^-11 for this class.

Edited by fancycat
Link to comment
Share on other sites

there is no change in mass. both objects have the same mass which is a constant.

Reduced mass.

This is because both objects are moving. You can either think of it as each object being attracted by a baricenter at reduced mass, or you can think of it as gravitational potential energy ending up being split between the two objects. Either way, you end up with velocity less by sqrt(2). And that's the correct answer.

Edited by K^2
Link to comment
Share on other sites

Reduced mass.

This is because both objects are moving. You can either think of it as each object being attracted by a baricenter at reduced mass, or you can think of it as gravitational potential energy ending up being split between the two objects. Either way, you end up with velocity less by sqrt(2). And that's the correct answer.

so we have to devide by two before taking the root. why isnt this in the solution on the key? so weird.

Edited by fancycat
Link to comment
Share on other sites

I don't know. It's pretty obvious to anyone who does orbital mechanics regularly. The moment I saw your post about two spheres I knew what was going on. Maybe whoever made the key just didn't think of it as something requiring explanation.

But yeah, I can see how this can be confusing if you aren't used to it. When you have just two objects, -GmM/R is the TOTAL gravitational energy. You don't get -GmM/R per object, but rather for the whole system. There is a way to derive that, of course, but it requires a bit of algebra. So GMM/a - GMM/b is the amount of energy that will be shared between the two neutron stars in this problem right before impact. That means the kinetic energy of each object in CoM frame is going to be M(GM/a - GM/b)/2, and this gives velocity as sqrt(GM/a - GM/b). The factor of two goes away.

You should also keep in mind the fact that the equations you are using are from classical mechanics. The answer you get is almost a tenth of the speed of light. That should be a warning sign that classical equations are probably going to be wrong. And indeed, when computing velocities for collision with neutron star, especially if the thing colliding into it is another neutron star, you have to use equations from General Relativity. To be honest, I'm not sure I'd know where to start on solving this problem correct. I know how to solve for a small object, like an asteroid, striking a neutron star, and I know the equations I'd use would alredy give me a different answer.

Link to comment
Share on other sites

This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...