VeritasVosLiberabit Posted July 7, 2012 Share Posted July 7, 2012 I have some questions regarding the gravitational constant and Newton's theory of gravity. To start, I am a first year physics undergrad student just so you have a sense of my caliber of understanding. Here is my question. As I understand Newton was able to come up with an equation for the gravitational force of attraction between any two massive bodies F= (M1*m2)/d^2 Later, Henry Cavendish invented an experiment in order to find the gravitational constant (6.673 * 10^-11 kg^-1*m^3*s^-2), originally intended to find the density of Earth, which happened a number of years after Newton's death. Cavendish's constant was about 1% off the number we use today. What I'm trying to understand is what is the purpose behind the constant or proportionality. Why is the gravitational constant even necessary, especially if Newton was already able to derive the original equation for the force of gravity between any two massive bodies without this constant? What I'm mostly looking for is a better understanding for constants of proportionality. mu_0 (permeability constant in magnetic field equation) is also an example of a constant I'm still having trouble understanding the purpose of. I simply am having trouble understanding why the main variables involved in these processes simply aren't enough to satisfy their respective equations, why is a proportionality constant so necessary? Link to comment Share on other sites More sharing options...

juanrga Posted July 7, 2012 Share Posted July 7, 2012 (edited) I have some questions regarding the gravitational constant and Newton's theory of gravity. To start, I am a first year physics undergrad student just so you have a sense of my caliber of understanding. Here is my question. As I understand Newton was able to come up with an equation for the gravitational force of attraction between any two massive bodies F= (M1*m2)/d^2 Later, Henry Cavendish invented an experiment in order to find the gravitational constant (6.673 * 10^-11 kg^-1*m^3*s^-2), originally intended to find the density of Earth, which happened a number of years after Newton's death. Cavendish's constant was about 1% off the number we use today. What I'm trying to understand is what is the purpose behind the constant or proportionality. Why is the gravitational constant even necessary, especially if Newton was already able to derive the original equation for the force of gravity between any two massive bodies without this constant? What I'm mostly looking for is a better understanding for constants of proportionality. mu_0 (permeability constant in magnetic field equation) is also an example of a constant I'm still having trouble understanding the purpose of. I simply am having trouble understanding why the main variables involved in these processes simply aren't enough to satisfy their respective equations, why is a proportionality constant so necessary? The force is |F| = G (m_{1}*m_{2})/d^{2} Apart from giving the correct units for the force, the constant G is also a measure of the strength of the gravitational interaction (G appears in other equations of physics). In a hypothetical universe where G takes twice its value here, the gravitational force would be more important than in our universe. Edited July 7, 2012 by juanrga 1 Link to comment Share on other sites More sharing options...

PSM5 Posted July 8, 2012 Share Posted July 8, 2012 I simply am having trouble understanding why the main variables involved in these processes simply aren't enough to satisfy their respective equations, why is a proportionality constant so necessary? It's not really necessary. It just makes the units come out right. If units are not important for what you're trying to show then you can just set it's value to 1 and leave it out. An example is this wikipedia page showing the proportional equivalence of the three types of mass: http://en.wikipedia.org/wiki/Equivalence_principle#Active.2C_passive.2C_and_inertial_masses As juanrga pointed out, there is another way to look at this. However, I'm not sure if this would be an academically correct way of looking at it. And that way is to consider the value of G to be a constant of proportionallity between active gravitational mass and inertial mass. For example, if two universes (A and B) were identical in every way except one had a higher or lower measured value of G than the other, then the gravitational force between two bodies in A and the gravitational force between two bodies in B would be different, even though the inertial masses of all four bodies were the same. In other words, G determines how strong the gravitaional field is for any particular unit of inertial mass. 1 Link to comment Share on other sites More sharing options...

pmb Posted July 8, 2012 Share Posted July 8, 2012 It's not really necessary. It just makes the units come out right. As juan noted, it also adjusts the value of the strength of the gravitational field And that way is to consider the value of G to be a constant of proportionallity between active gravitational mass and inertial mass. If a body A exerts a force on body B then its the active gravitational mass which is acting by A and its the passive gravitational mass that is being acted on B. Therefore we let M = active gravitational mass of body A m = passive gravitational mass of bdy B The force on B due to A is then F = GMm/r^{2} The gravitational constant G then determines the strength of the force. M and m are found to be equal to the inertial mass of body A and B respectively. Link to comment Share on other sites More sharing options...

PSM5 Posted July 8, 2012 Share Posted July 8, 2012 As juan noted, it also adjusts the value of the strength of the gravitational field Yes, I was agreeing with him on that point. If a body A exerts a force on body B then its the active gravitational mass which is acting by A and its the passive gravitational mass that is being acted on B. Therefore we let M = active gravitational mass of body A m = passive gravitational mass of bdy B The force on B due to A is then F = GMm/r^{2} The gravitational constant G then determines the strength of the force. M and m are found to be equal to the inertial mass of body A and B respectively. I agree with your post. However, I would just like to add that the use of the term passive gravitational mass in the universal law of gravitation is just a convention. It is not necessary. The split of gravitational mass into it's current two forms, active and passive, did not even exist before 1957. For example, when you apply Newton's second law to the universal law of gravitation to get the acceleration of m_{p}, you are canceling out the passive gravitational mass m_{p} in the universal law of gravitation with the inertial mass m_{i} in Newton's second law. If they were not the same then you should not be able to do this. Having three different types of mass can be quite confusing. Link to comment Share on other sites More sharing options...

pmb Posted July 8, 2012 Share Posted July 8, 2012 Having three different types of mass can be quite confusing. I quite agree. However I've found it useful to keep them in mind since the active gravitational mass densty is not equal to passive gravitational mass density. Link to comment Share on other sites More sharing options...

elfmotat Posted July 8, 2012 Share Posted July 8, 2012 Here's how I'd think about it: the units we use are rather arbitrary. Humans (originally) defined the kilogram, the second, and the meter in terms of familiar mass/time/length scales. Since Newtons (force) are derived from those units, the Newton is also a rather arbitrary unit. The Law of Gravitation therefore pretty much requires a constant of proportionality in order to scale length, time, and mass into our arbitrary units. Planck units are much more fundamental: working in terms of Planck units means setting the speed of light, Planck's constant, and the gravitational constant all to one. Length, time, and mass units are defined in terms of fundamental constants. So, working in terms of Planck units, you can indeed say that simply F=m_{1}m_{2}/r^{2}. 1 Link to comment Share on other sites More sharing options...

michel123456 Posted July 8, 2012 Share Posted July 8, 2012 Here's how I'd think about it: the units we use are rather arbitrary. Humans (originally) defined the kilogram, the second, and the meter in terms of familiar mass/time/length scales. Since Newtons (force) are derived from those units, the Newton is also a rather arbitrary unit. The Law of Gravitation therefore pretty much requires a constant of proportionality in order to scale length, time, and mass into our arbitrary units. Planck units are much more fundamental: working in terms of Planck units means setting the speed of light, Planck's constant, and the gravitational constant all to one. Length, time, and mass units are defined in terms of fundamental constants. So, working in terms of Planck units, you can indeed say that simply F=m_{1}m_{2}/r^{2}. But G has units. You can't say it is equal to 1. Link to comment Share on other sites More sharing options...

elfmotat Posted July 8, 2012 Share Posted July 8, 2012 But G has units. You can't say it is equal to 1. Uh... yes you can. Link to comment Share on other sites More sharing options...

pmb Posted July 8, 2012 Share Posted July 8, 2012 (edited) Uh... yes you can. You're right. I know that in general relativity one tends to use God's unts. If M and m have units of klilogams [kg] and r has units of meters [m] then Mm/r^{2} has units of [kg]^{2}[m]^{-2} which isn't the units of force, which is Newons [N] = [kg][m]^{-2} which is wrong, as is the magnitude. What one needs to do is to change the units of distance and mass according to Plnak's units See http://en.wikipedia.org/wiki/Planck_units I find those units cumbersome. I prefer to use in MKS units so I can easily compute values of the gravitational force. Edited July 8, 2012 by pmb Link to comment Share on other sites More sharing options...

michel123456 Posted July 9, 2012 Share Posted July 9, 2012 Uh... yes you can. You're right. I know that in general relativity one tends to use God's unts. If M and m have units of klilogams [kg] and r has units of meters [m] then Mm/r^{2} has units of [kg]^{2}[m]^{-2} which isn't the units of force, which is Newons [N] = [kg][m]^{-2} which is wrong, as is the magnitude. What one needs to do is to change the units of distance and mass according to Plnak's unitsSee http://en.wikipedia.org/wiki/Planck_units I find those units cumbersome. I prefer to use in MKS units so I can easily compute values of the gravitational force. but Planck units can be translated in SI units: Planck length=1.616 199(97) × 10−35 m Planck mass=2.176 51(13) × 10−8 kg Planck time=5.391 06(32) × 10−44 s You cannot "change the units of distance and mass according to Planck's units" and obtain G=1. The only thing you can do is erase the units completely and retain the numerical values. When you insert proper units, dimensional analysis shows that G must keep units. G=1 is only a mathematical trick. Link to comment Share on other sites More sharing options...

swansont Posted July 9, 2012 Share Posted July 9, 2012 but Planck units can be translated in SI units: Planck length=1.616 199(97) × 10−35 m Planck mass=2.176 51(13) × 10−8 kg Planck time=5.391 06(32) × 10−44 s You cannot "change the units of distance and mass according to Planck's units" and obtain G=1. The only thing you can do is erase the units completely and retain the numerical values. When you insert proper units, dimensional analysis shows that G must keep units. G=1 is only a mathematical trick. Yes, it's a trick. But theoreticians often find it to be a useful trick. http://en.wikipedia.org/wiki/Natural_units Link to comment Share on other sites More sharing options...

rizal23 Posted July 24, 2012 Share Posted July 24, 2012 (edited) I have some questions regarding the gravitational constant and Newton's theory of gravity. To start, I am a first year physics undergrad student just so you have a sense of my caliber of understanding. Here is my question. As I understand Newton was able to come up with an equation for the gravitational force of attraction between any two massive bodies F= (M1*m2)/d^2 Later, Henry Cavendish invented an experiment in order to find the gravitational constant (6.673 * 10^-11 kg^-1*m^3*s^-2), originally intended to find the density of Earth, which happened a number of years after Newton's death. Cavendish's constant was about 1% off the number we use today. What I'm trying to understand is what is the purpose behind the constant or proportionality. Why is the gravitational constant even necessary, especially if Newton was already able to derive the original equation for the force of gravity between any two massive bodies without this constant? What I'm mostly looking for is a better understanding for constants of proportionality. mu_0 (permeability constant in magnetic field equation) is also an example of a constant I'm still having trouble understanding the purpose of. I simply am having trouble understanding why the main variables involved in these processes simply aren't enough to satisfy their respective equations, why is a proportionality constant so necessary? Formula of the gravitational force is F = (G * M_{1} * m_{2})/R^{2} info: G : gravitational constant M_{1} : active gravitational mass of body 1 m_{2} : passive gravitational mass of body 2 R : the radius of the object vBulletin Community Forum Edited July 24, 2012 by rizal23 Link to comment Share on other sites More sharing options...

alpha2cen Posted July 24, 2012 Share Posted July 24, 2012 (edited) The force is |F| = G (m_{1}*m_{2})/d^{2} Apart from giving the correct units for the force, the constant G is also a measure of the strength of the gravitational interaction (G appears in other equations of physics). In a hypothetical universe where G takes twice its value here, the gravitational force would be more important than in our universe. Is Gravity constant linear to the amount of mass? For example A mass 1 object........................................ B gravity A mass 2 objects......................................2B gravity .. .. A mass 1*10^{100000} objects....... 1*10^{100000} B gravity Two problems. 1mass is A kg, total units is 1*10^{100000 }. Total mass is 1*10^{100000}A? Total Gravity is 1*10^{100000}B? Edited July 24, 2012 by alpha2cen Link to comment Share on other sites More sharing options...

juanrga Posted July 24, 2012 Share Posted July 24, 2012 (edited) Is Gravity constant linear to the amount of mass? No sure what do you mean by "linear", but notice that a constant does not vary with "the amount of mass", otherwise it cannot be a constant. Edited July 24, 2012 by juanrga Link to comment Share on other sites More sharing options...

alpha2cen Posted July 24, 2012 Share Posted July 24, 2012 (edited) No sure what do you mean by "linear", but notice that a constant does not vary with "the amount of mass", otherwise it cannot be a constant. F_{g}=G^{o}m, or F_{g}=G^{'}m where G^{'}=G^{o}+bm Is the constant b always 0 no matter how big the mass size is? Edited July 24, 2012 by alpha2cen Link to comment Share on other sites More sharing options...

swansont Posted July 24, 2012 Share Posted July 24, 2012 F_{g}=G^{o}m, or F_{g}=G^{'}m where G^{'}=G^{o}+bm Is the constant b always 0 no matter how big the mass size is? |F| = G Mm/r^2 Yes, it's linear with mass, insofar as Newtonian gravity is valid. Link to comment Share on other sites More sharing options...

alpha2cen Posted July 24, 2012 Share Posted July 24, 2012 Does Gravity constant have the same value anywhere in the Universe? How do we know it? Link to comment Share on other sites More sharing options...

juanrga Posted July 24, 2012 Share Posted July 24, 2012 F_{g}=G^{o}m, or F_{g}=G^{'}m where G^{'}=G^{o}+bm Is the constant b always 0 no matter how big the mass size is? I suppose that by G^{'} you mean the gravitational constant G. In that case, yes, b is zero. Otherwise G could not be a constant! If by F_{g} you mean gravitational force, the expression that you give is only valid for small masses and velocities. Link to comment Share on other sites More sharing options...

burhan hafiz Posted July 25, 2012 Share Posted July 25, 2012 gravitaitional constant is necessary in the equation otherwise the magnitude of force changes. Link to comment Share on other sites More sharing options...

VeritasVosLiberabit Posted July 29, 2012 Author Share Posted July 29, 2012 (edited) So I have really been pondering this question for the past few weeks to find the relationship between this constant 'G' and its relationship in the equation for the force of gravity between two massive bodies. The equation for a proportionality relationship is y = kx. This formula looks familiar because Hooke's law is nearly identical to this except 'k' is made negative (not sure why but perhaps because spring rest is 0 in cartesian plane and spring force direction opposite of x?). Anyway, 'k' represents a constant and a constant is a variable used to show a relationship between two or more coordinates. In our case with gravity, we want to see how the force of gravity increases or decreases between any two massive bodies. The two factors that have significance in changing our gravitational force are distance and mass. *The strength of the force of gravity changes at a ratio of 6.673 * 10^-11 for every kilogram of mass with respect to the distance in meters (squared). Basically, this gravitational constant tells us by exactly what interval the force of gravity between these two massive bodies changes. F = mg is just like the equation y = kx. Its the equation for proportionality between two coordinates. this formula is simplified because g = GM/r^2. I'm still not very sure how Newton could have derived the formula F = (M1 * m2) / r^2 without the gravitational constant, because to me it still seems you would need a constant of proportionality in order to calculate exactly by what factor the mass and distance change the force of gravity between the two bodies. It seems that without that number you can understand the relationship among mass and distance and the force of gravity but no actual calculations could be obtained without a constant. Perhaps this could be due to my currently low understanding of different units systems. Some final thoughts I had on the force of gravity: I started wondering why we don't use an intrinsic gravity number in which we can say this planet has a mass of 'x' kg therefore it has an intrinsic gravitational pull of 'x' newtons, similarly to voltage in a battery. But then I remembered that a potential difference, which is the potential energy between two points, constitutes the number representing voltage for a battery and the current is more important than the voltage anyway. But I came to the conclusion that maybe we don't have this intrinsic number because mass in itself is already somewhat of an "intrinsic "gravitational number." Mass tells you that if the number is really high there will be lots of gravity. So gravity is a number that will tell you the influence it has on a specific object which is incredibly useful. I hope what I have written was clear enough to understand my interpretations on the variables associated with the formula for gravitational force. * I put an asterix by this sentence because this phrase here brought up a inquiry for me. How can we set this constant to m^3/(s^2kg) just to get the units to cancel? Why should the units need extra help to get cancelling from what should be a unitless ratio? Edited July 29, 2012 by VeritasVosLiberabit Link to comment Share on other sites More sharing options...

swansont Posted July 29, 2012 Share Posted July 29, 2012 So I have really been pondering this question for the past few weeks to find the relationship between this constant 'G' and its relationship in the equation for the force of gravity between two massive bodies. The equation for a proportionality relationship is y = kx. This formula looks familiar because Hooke's law is nearly identical to this except 'k' is made negative (not sure why but perhaps because spring rest is 0 in cartesian plane and spring force direction opposite of x?). It's negative because the force is attractive for x>0. I'm still not very sure how Newton could have derived the formula F = (M1 * m2) / r^2 without the gravitational constant, because to me it still seems you would need a constant of proportionality in order to calculate exactly by what factor the mass and distance change the force of gravity between the two bodies. Look at ratios of forces. G will cancel. Link to comment Share on other sites More sharing options...

PSM5 Posted July 29, 2012 Share Posted July 29, 2012 I started wondering why we don't use an intrinsic gravity number in which we can say this planet has a mass of 'x' kg therefore it has an intrinsic gravitational pull of 'x' newtons, similarly to voltage in a battery. Are you familiar with the standard gravitational parameter? It doesn't have units of force, but it is a direct measure of a body's "gravitational charge". Link to comment Share on other sites More sharing options...

VeritasVosLiberabit Posted July 29, 2012 Author Share Posted July 29, 2012 It's negative because the force is attractive for x>0. Look at ratios of forces. G will cancel. ah I see now. y = kx -> F = G( m*M/r^2) y = F = N = (kgm)/s^2 x = (m*M)/ r^2 = kg^2/m^2 k = y/x -> G = [(kgm)/s^2]/(kg^2/m^2) = [(kgm)/s^2](m^2/kg^2) = m^3/s^2kg or m^3 kg^-1 s^-2 should have done this mathematically from the beginning instead of trying to use intuition, thanks for pointing this out. Link to comment Share on other sites More sharing options...

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