geko Posted September 8, 2007 Report Posted September 8, 2007 I have this problem with a question i need to do. The whole question is Calculate how many times stronger the electric force is than the gravitaional force, between a proton of charge +e and mass Mp at a distance r from an electron of charge -e and mass Me given that: e = 1.6 x 10 to -19 CMe = 9.1 x 10 to -31 kgMp = 1.67 x 10 to -27 kgKe = 9.0 x 10 to 9 N M CG = 6.7 x 10 to -11 N m kg My problem is that the only 2 equations i can think of using for it are coulombs law and newtons law of gravity. The trouble is i dont know r, so i cant simply calculate them out. There's also no point i think in rearranging the formulas for r to find it since i dont know F in either case. Totally stuck on this, is there another equation that i may have overlooked that i could use? Or maybe something else im overlooking? Any help appreciated.
modest Posted October 4, 2007 Report Posted October 4, 2007 I have this problem with a question i need to do. The whole question is Calculate how many times stronger the electric force is than the gravitaional force, between a proton of charge +e and mass Mp at a distance r from an electron of charge -e and mass Me given that: e = 1.6 x 10 to -19 CMe = 9.1 x 10 to -31 kgMp = 1.67 x 10 to -27 kgKe = 9.0 x 10 to 9 N M CG = 6.7 x 10 to -11 N m kg My problem is that the only 2 equations i can think of using for it are coulombs law and newtons law of gravity. The trouble is i dont know r, so i cant simply calculate them out. There's also no point i think in rearranging the formulas for r to find it since i dont know F in either case. Totally stuck on this, is there another equation that i may have overlooked that i could use? Or maybe something else im overlooking? Any help appreciated. You could pick an arbitrary distance (i.e. r=1) for both equations and solve for force or you could solve for a constant in the equation setting them equal: [math] \text{constantX =}\left[\frac{K_c q_1 q_2}{G M_a M_b}\right] [/math] where constantX is the number of times the electrostatic force is larger than the force of gravityKc is the electrostatic constant 8.988 x 10^9both q1 and q2 are positive absolute values 1.6 x 10^-19 And, you should get the electrostatic force being around 10^39 times stronger-modest
heterotic Posted July 8, 2010 Report Posted July 8, 2010 You cannot set r to an arbitrary number, because those forces are not proportional to each other. Your problem says you are dealing with a proton and an electron, so you need to calculate the orbital radius of the electron in a Hydrogen atom, with the standard e+/e-, the electron is at ground state (n = 1). So, you can use the Bohr Radius. Here's how you get r, starting with de Broglie's hypothesis. λ(n) = 2πr/n This predicts a momentum (Bohr's hypothesis) rp = nħ p = nħ/r so mvr = nħ So for n = 1 λ = 2πr mvr = ħ p = ħ/r So we look for the Electric Potential Energy, where ε = 8.854188e-12 F/m (the electric constant in a vacuum) E = (1/4πε)(Q1Q2/r) Substitute e for Q1 and Q2 E = (1/4πε)(e^2/r) If we apply Newton's Laws m(v^2/r) = p^2/mr = (1/4πε)(e^2/r^2) Rearrange for p p^2 = (1/4πε)(me^2/r) Substitute our deBroglie/Bohr equations (ħ/r)^2 = (1/4πε)(me^2/r) Shuffle that for r r = 4πεħ^2/me^2 or r = ħ^2/mcα α = Ke^2/ħc so r = ħ^2/mKe^2
Pyrotex Posted July 8, 2010 Report Posted July 8, 2010 ...trouble is, I don't know r... I googled "size hydrogen atom" -- and the very first link was to a page that stated the distance ® between the proton and electron in a hydrogen atom was .000048 micron. a micron is 10^-6 meter. So r is approximately 5 * 10^-15 meter.
heterotic Posted July 8, 2010 Report Posted July 8, 2010 I googled "size hydrogen atom" -- and the very first link was to a page that stated the distance ® between the proton and electron in a hydrogen atom was .000048 micron. a micron is 10^-6 meter. So r is approximately 5 * 10^-15 meter. What if this question was on an exam? You can't just google the answers. Unless you are given the Bohr Radius (or know it), you have to calculate it. On an exam, even if I have something memorized, like the Bohr Radius, I still show how to calculate it, unless it is given by the professor. Giving someone the answer is not properly helping them, especially when it is incorrect. The radius, using the correct calculation listed above is 0.529e-10 m, or about .529 Angstrom.
modest Posted July 10, 2010 Report Posted July 10, 2010 You cannot set r to an arbitrary number, because those forces are not proportional to each other. Sure they are. They are both inversely proportional to the square of the distance. The answer depends on the charge of the particle and the mass of the particle, but in no way depends on the distance between them. The answer to the question is X in the following equation: [math]K_e\frac{q_1q_2}{r^2}=G\frac{M_1M_2}{r^2}X[/math] The left hand side is the electrostatic force and the right hand side is the gravitational force where X is a proportionality constant. The r's cancel (the answer doesn't depend on distance). So, you don't need the size of a hydrogen atom. To prove this to yourself, solve the forces using the numbers from the opening post for different values of r:[math]F_G=G \frac{M_1 M_2}{r^2}[/math][math]F_C=K_e \frac{q_1 q_2}{r^2}[/math]Solve these two equations for 1 meter or for Pyro's 5 x 10-15 meters. Either way FC will be about 10^39 times larger than FG. ~modest
heterotic Posted July 10, 2010 Report Posted July 10, 2010 Sure they are. They are both inversely proportional to the square of the distance. The answer depends on the charge of the particle and the mass of the particle, but in no way depends on the distance between them. The answer to the question is X in the following equation: [math]K_e\frac{q_1q_2}{r^2}=G\frac{M_1M_2}{r^2}X[/math] The left hand side is the electrostatic force and the right hand side is the gravitational force where X is a proportionality constant. The r's cancel (the answer doesn't depend on distance). So, you don't need the size of a hydrogen atom. To prove this to yourself, solve the forces using the numbers from the opening post for different values of r:[math]F_G=G \frac{M_1 M_2}{r^2}[/math][math]F_C=K_e \frac{q_1 q_2}{r^2}[/math]Solve these two equations for 1 meter or for Pyro's 5 x 10-15 meters. Either way FC will be about 10^39 times larger than FG. ~modest You're right. The force of electricity will be 2.27e39 times greater than the force of gravity no matter the distance you give in that single equation you've given, with the given information. That wasn't my point. The mass of a particle is variable. Given the invariant mass, that means the system is only an electron. This is one of those things where you don't need the radius to solve the math problem, but to apply it to physics, it doesn't make sense to say that the radius doesn't matter. If you want to use that problem, and always get 2.27e39 you have to use r > 5ħ^2/mKe^2 (a free electron and a free proton) Anything less than or equal to 5ħ^2/mKe^2 would be a Hydrogen atom, and would change the masses of the two.
modest Posted July 11, 2010 Report Posted July 11, 2010 You're right. The force of electricity will be 2.27e39 times greater than the force of gravity no matter the distance you give in that single equation you've given, with the given information. I didn't give the equations. The OP did. And, yes, I agree—the distance is irrelevant. ~modest
heterotic Posted July 11, 2010 Report Posted July 11, 2010 I didn't give the equations. The OP did. And, yes, I agree—the distance is irrelevant. ~modest Actually, you gave him the proportionality equation above. The distance is irrelevant mathematically, but not physically.
modest Posted July 11, 2010 Report Posted July 11, 2010 If you think it's physically important, you could solve the binding energy of a hydrogen atom for us. Given the mass of an electron and proton, what is the expected mass of a hydrogen atom? ~modest
heterotic Posted July 11, 2010 Report Posted July 11, 2010 E(k) = E(0) - hcR(Z^2/n^2) E(k) = m(0)c^2 - hcR(Z^2/n^2) E(k) = (9.10938215e-31 * 299792358^2) - (6.626e-34 * 299792458 * 1.097373e7(1^2/1^2)) E(k) = 8.187104e-14 - 2.179848971359038084e-18 E(k) = 8.186886e-14 J m = E(k)/c^2 m(e) = 8.186886e-14/299792358^2 = 9.10914524377e-31 kg m(p) = m(H) - m(e) m(p) = (1.0794/6.0221415e23)/1000 - 9.10914524377e-31 m(p) = 1.79238564885e-27 - 9.10914524377e-31 = 1.792385648e-27 kg F© = 8.98755e9(1.603e-19^2/0.529e-10^2) = 8.25272e-8 N F(g) = 6.67428e-11((9.10914524377e-31*1.792385648e-27)/0.529e-10^2)) = 3.89406e-47 N The Force of Electricity is 4.719e40 times stronger than Force of Gravity at this radius. NOT 2.27e39
modest Posted July 11, 2010 Report Posted July 11, 2010 ...m(H)... = (1.0794/6.0221415e23)/1000... So, you googled it. You didn't solve 1.0794, nor is Avogadro's number given in the OP. Not to mention, the extra mass of a hydrogen atom comes from the virtual photons (and the gluons if you're using 1.0794 which has an isotope greater than 1), not so much an increase in mass of the proton and electron. ~modest
heterotic Posted July 11, 2010 Report Posted July 11, 2010 So, you googled it. You didn't solve 1.0794, nor is Avogadro's number given in the OP. Not to mention, the extra mass of a hydrogen atom comes from the virtual photons (and the gluons if you're using 1.0794 which has an isotope greater than 1), not so much an increase in mass of the proton and electron. ~modest I didn't google anything. I know the mass of a mole of H-1 (we figured that much out in AP Chemistry in high school, maybe your school left you only with Google for Chemistry), and Avogadro's number is a constant, it doesn't need to be given. Even if I had used something I didn't have memorized, like the mass of a mole of H-3, doesn't anyone remember there are books?! Looks like I should have looked it up. The AMDC says it is 1.00782503207(10) u, so I guess it's been updated. The excess mass of H-1 you are talking about is not an actual excess, because we cannot as of yet actually measure particles. It is just more than what we would "expect". Considering the fact that we are now diving into theoretical particle physics, neither one of us can be correct in measuring the mass of the proton, because you believe that a virtual photon is existing (not just mathematically) in the atom to carry the electromagnetic force, (I'm disregarding your comment about gluons, since they are binding quarks, which would be what makes up the proton), but now we have to consider a virtual particle also carrying the gravitational force. That really doesn't change the fact that the masses of the particles are variable. If you really want to get down into virtual particles, fine, but you're only disproving your point about the radius being irrelevant. In certain orbitals, the electron self-interacts, causing the electric forces between it and the proton to decrease slightly. This has nothing to do with the OP's problem anymore, I'm making a point to you, and you saying that the radius is irrelevant, when it's really not. Mass changes in proportion to kinetic energy, does it not? The kinetic energy of these particles is variable, so their relativistic masses are as well.
modest Posted July 11, 2010 Report Posted July 11, 2010 I'm sorry, heterotic, but your objection just wasn't relevant to the problem. The answer to the OP is in the post directly following the OP. ~modest
heterotic Posted July 11, 2010 Report Posted July 11, 2010 I'm sorry, heterotic, but your objection just wasn't relevant to the problem. The answer to the OP is in the post directly following the OP. ~modest Actually, it depends on what his purpose is and what level physics he's in. None of my professors would have accepted "10^39 times stronger", since with the mathematics of the given information is 2.27 x 10^39. Furthermore, none of my physics professors would have accepted that answer period, for the objection I stated, unless you point out that radius has to be great enough that the two are not contained within a system that becomes an atom to get a constant proportion. If he's doing for reasons that have nothing to do with school (a fair assumption given the fact that it is July), then it really matters, because it's more than a math problem. I'm glad you edited your post, though from "In any case, the answer to the OP is in the post directly following the OP.", since the answer you give the OP is wrong.
modest Posted July 11, 2010 Report Posted July 11, 2010 You are dissembling. The question of the OP "My problem is that the only 2 equations i can think of using for it are coulombs law and newtons law of gravity. The trouble is i dont know r" is that r is irrelevant. Any r you choose will give you the same answer. Your objection "You cannot set r to an arbitrary number, because those forces are not proportional to each other" is flat out wrong. They are precisely proportional. ~modest
heterotic Posted July 11, 2010 Report Posted July 11, 2010 You are dissembling. The question of the OP "My problem is that the only 2 equations i can think of using for it are coulombs law and newtons law of gravity. The trouble is i dont know r" is that r is irrelevant. Any r you choose will give you the same answer. Your objection "You cannot set r to an arbitrary number, because those forces are not proportional to each other" is flat out wrong. They are precisely proportional. ~modest As I said before, you're right, they are proportional as long as you ignore the physics in the calculation. If you're doing nothing other than solving a math problem to prove that the strength of the electric force is stronger than the gravitational force between a proton and an electron, it doesn't matter if the answer is accurate. If you're using that information to prove a theory, using further proofs, I suggest you take into account the natural order of these particles.
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