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A Physics or Math Question
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LegendaryDreams
Canada1350 Posts
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fanatacist
10319 Posts
g = gravitation constant m = mass 1 M = mass 2 r = distance from centers of mass (Google the gravitation constant, I think it is something like 6.637 x 10^-11 but I haven't used it in a while so I may be mistaken)\ EDIT: Actually, you don't need the constant, since it's just a ratio. If you want me to do it for you, respond I guess. | ||
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LegendaryDreams
Canada1350 Posts
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BottleAbuser
Korea (South)1888 Posts
I'm going to arbitrarily call the mass of Earth x, the mass of the moon y, and the mass of the sun y*3*10^7. The distance between Earth and moon is 1 in arbitrary units, and distance between Earth and sun is 400. (All givens.) Now, given equation G = g * mM*r^-2 z * G(moon) = G(sun), and we're solving for z. Plugging in values for g, m, M, and r for each side of the equation: xyzg = xyg *3*10^7 * 400^-2 Simplifying: z = 3 * 10 ^ 7 * 160000 ^ -1 = 3 * 10 ^ 2 / 1.6 which is roughly z = 200 So, the pull on Earth by the sun is about 200 times stronger than the pull on Earth by the moon. Unless I've dropped a zero somewhere or a sign. Which would be embarrassing as hell. | ||
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BottleAbuser
Korea (South)1888 Posts
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eVo[LvE]
Canada19 Posts
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BottleAbuser
Korea (South)1888 Posts
Maybe you ask OP because you've seen same problem as hw?  | ||
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fanatacist
10319 Posts
On January 29 2008 10:07 BottleAbuser wrote: Er, yeah. I'm bored, so everyone reading this will have to see my mess. I'm going to arbitrarily call the mass of Earth x, the mass of the moon y, and the mass of the sun y*3*10^7. The distance between Earth and moon is 1 in arbitrary units, and distance between Earth and sun is 400. (All givens.) Now, given equation G = g * mM*r^-2 z * G(moon) = G(sun), and we're solving for z. Plugging in values for g, m, M, and r for each side of the equation: xyzg = xyg *3*10^7 * 400^-2 Simplifying: z = 3 * 10 ^ 7 * 160000 ^ -1 = 3 * 10 ^ 2 / 1.6 which is roughly z = 200 So, the pull on Earth by the sun is about 200 times stronger than the pull on Earth by the moon. Unless I've dropped a zero somewhere or a sign. Which would be embarrassing as hell. Uh... I think that's wrong. Here is the solution I made, maybe you can find a mistake: Well, GmM/R^2 is an old formula for gravitation and orbitting of planets. It measures the strength of gravitation between two objects, so it is perfect in this context. Here is how I would solve it: m1 = Sun m2 = Moon M = Earth R1 = Distance from Sun to Earth R2 = Distance from Moon to Earth Gm1M/R1^2 = Strength of gravitation of Sun to Earth Gm2M/R2^2 = Strength of gravitation of Moon to Earth So, to answer your question, we simply plug in the data we know. Sun = 30 million times more massive than the Moon, so m1 = 3 * 10^7 * m2 Sun = 400 times further away than the Moon, so R1 = 4 * 10^2 * R2 So, now we simply plug in the data back into the original formula, and compare them as a ratio Strength of Gravitation Sun : Moon Gm1M/R1^2 : Gm2M/R2^2 G(3 * 10^7 * m2)M/(4 * 10^2 * R2)^2 : Gm2M/R2^2 Here we can already see that you can divide both sides by G and M. This leaves us with: (3 * 10^7 * m2)/(4 * 10^2 * R2)^2 : m2/R2^2 Now we can divide both sides by m2 and multiply by R2^2, to have a pure number ratio: (3 * 10^7)/(4 * 10^2)^2 : 1/1 (3 * 10^7)/(16 * 10^4) : 1 (3/16)(10^3) : 1 187.5 : 1 Obviously, the Sun's force of gravity on the Earth is much greater than the Moon's. In fact, it is 187.5 times stronger! I hope you understood my broken English, I can try to clarify parts n_n. Although, like you said, you might have fucked up. I think my work is neater and in more steps. EDIT: HAHA forgot to square R. Hold on. EDIT 2: Fixed. I fucking suck. My defense is that I did this 3 years ago T_T. | ||
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BottleAbuser
Korea (South)1888 Posts
The givens are to 1 significant digit (distance ratio is closer to 390, mass ratio is closer to 2.7*10^7). Answer should also be to 1 significant digit. | ||
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fanatacist
10319 Posts
On January 29 2008 10:39 BottleAbuser wrote: Yay for nitpicking. The givens are to 1 significant digit (distance ratio is closer to 390, mass ratio is closer to 2.7*10^7). Answer should also be to 1 significant digit. This level of physics is barely an accurate measure worthy of significant figures, I believe, because GmM/R^2 basically assumes that the planets are spherical point-masses, so it doesn't hurt to ignore sig figs in this, just like you would in a math problem. At least that's the way I was taught this relatively low-difficulty conceptual physics stuff. It's not like read-outs from chemistry conversions and experiments xD. | ||
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BottleAbuser
Korea (South)1888 Posts
It's valid to treat them as spheres (and spheres are equivalent to point masses in this model), because they're very damn similar to perfect spheres, and the composition is very uniform among any given shell within those spheres. It's always a good thing to observe significant figures! In pure mathematics, there is no uncertainty (even with probability your level of uncertainty is very certain), so there is no need for significant figures. When we're talking about models of the real world (be it chemical reactions or bigger objects), we're using measurements as input, and those always have some level of uncertainty... and significant figures is how we deal with the uncertainty! | ||
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LxRogue
United States1415 Posts
On January 29 2008 09:49 LegendaryDreams wrote: The nearest decent-sized celestial body to the earth is the moon. But we don`t orbit about that. To see why we orbit about the sun, rather than orbit about the moon, work out how much stronger the force of gravity is from the sun than the force of gravity from the moon. All you need to know is that the sun is about 30 million (3x10 to the 7) times as massive as the moon, and about 400 times further away. You don't need any constants or stuff, you just compare the two forces with the inverse square law. Gm1m2/r^2 = Gm1m2/r^2 G and m1 (mass of the earth) cancel out and you know the relationship between the distances and m2. 30,000,000 * m2 / (400r)^2 = m2 / r^2 so the force due to the sun is 30,000,000 / 400^2 times as strong or 187.5 times as strong as the moons force. (edit: whoops i guess im the 3rd person to get this) | ||
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fanatacist
10319 Posts
On January 29 2008 10:56 BottleAbuser wrote: Nitpick wars begin! It's valid to treat them as spheres (and spheres are equivalent to point masses in this model), because they're very damn similar to perfect spheres, and the composition is very uniform among any given shell within those spheres. It's always a good thing to observe significant figures! In pure mathematics, there is no uncertainty (even with probability your level of uncertainty is very certain), so there is no need for significant figures. When we're talking about models of the real world (be it chemical reactions or bigger objects), we're using measurements as input, and those always have some level of uncertainty... and significant figures is how we deal with the uncertainty! Close... But not quite it. I think it's inferred that the answer 187.5 isn't meant to be taken as an accurate value - it's the mathematical result to a mathematical problem. It's obviously not exact, just like GmM/R^2 may be close but not exact either. If we wish to be technical, then yes in this case, as a physics problem, we would use sig figs. But as a math problem, (title = Math/Physics problem), that is the exact answer. You can implement sig figs at the end depending on what the problem is looking for, so isn't it better to give the exact mathematical answer and then let the OP decide which is more appropriate? Also, the OP's issue was more conceptual than mathematical, so either way I would not have cared about finer details such as sig figs, which I personally feel are unnecessary outside of finalized lab reports or final solutions to problems. This was just the methodology to solving the problem, and the exact mathematical result. I treated it as a math problem, ratios and all, independent of the real-world correlation. Didn't see much of a point in being so detailed about it  . | ||
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Luddite
United States2315 Posts
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eVo[LvE]
Canada19 Posts
On January 29 2008 10:18 BottleAbuser wrote: o.o CS at a Korean school. Maybe you ask OP because you've seen same problem as hw? Yeah I meant the OP lol. Thats the exact question on an assignment due this week :D | ||
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BottleAbuser
Korea (South)1888 Posts
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fanatacist
10319 Posts
On January 29 2008 14:42 Luddite wrote: To be fair, the earth does orbit around the center of the earth-moon system. A little bit. And every mass has a force on every other object in the universe xD. It's not nearly strong enough in comparison to the sun's gravitational force. The Earth wobbles a little bit, at most. | ||
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