Understanding the Inverse Square Law in Planetary Motion

In summary: The gravitational force is the force of gravity, which is the force of an object's mass attracting other objects with a gravitational force. When you divide both sides by the mass of the planet, the result is the centripetal acceleration. This is the acceleration that the planet experiences as it revolves around the sun.
  • #1
matheson
13
0
hey guys,
im doing an investigation into the accuracy of the "inverse square law" (any two particles with experience a mutually attractive force yadda yadda yadda)
anyway, we have to find various informaiton about the planents. (orbital velocities, centripetal accelerations, etc.)

my problem is: when calculating centripetal acceleration (using A = v^2/r), i got really small values, which doesn't seem right.

in the end, i couldn't get the inverse square law to match up (by comparing distances form the sun and using that ratio, inverting and squaring it, you should end up with the force of the planet having the ratio applied to it, if that makes sense).

SO basically id just like to know what the centripetal acceleration should be ( i used average values for radius and velocity, seing as planets revolve in an ecliptic manner), or if those values seem right, why doesn't the inverse square law match up with the calculated forces?
 
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  • #2
Your equation for centripetal acceleration is correct. The eliptic orbits are close enough to being circular so as not to cause significant difficulty in your calculations. Perhaps if you posted your calculations for one or two planets, we might see where things are going wrong.
 
  • #3
alright, these are done by hand so it might look a little messy:
start with mercury
r = 57900000000m } found on internet
m = 3.3 x 10^23

now since v = (square root) Gm/r [sub in values... m = mass of sun]
v= 47927.7m/s

since a = v^2/r = [sub in values]
a = 0.04m/s^2

since F=ma = [sub in values, m = mass of planet not sun ]
F = 1.32 x 10^22

from there you can compare the force and radius of mercury to that of another planet.
Assuming all above working is correct, so will calculations for Venus.
eg
r = 108200000000m
F = 5.36 x 10^22

then you can compare the radius's and come up with a ratio of about 1:1.87 for radius's of Mercury:Venus

so we should then be able to say that (using F proportional to 1/r^2) the force of Venus should be 1/(1.87^2) the force of Mercury, correct?
so it should be 3.77 x 10^21, but with my calculations it turned out being 5.36 x 10^22, which is quite different

thanks to all your help, this investigation is no doubt tame by your guys standards!
 
  • #4
What does it mean for an object to have a force? If there was no force would the object move?...and in what direction would the object move?

Is this the equation you are thinking of with the inverse square:

Universal Gravitation
[tex] F = \frac{Gm_1m_2}{r^2} [/tex]
where:
[tex] G = 6.67\times 10^{-11}Nm^2/kg^2 [/tex]
 
  • #5
i also used the above equation and the forces work out extemelly similar to the F=ma equaiotn (only different subject to rounding).

and yes that is the the equation it refers to, as in if the radius doubles, the force will quarter.
since I am not considering the masses however, can't it just be written as F (alpha) 1/r^2?
and so then shouldn't i be able to use the ratio system above?

please point out any errors, i can't see what I've done wrong, besides maybe rounding badly. has anyone tried the calculations to see it theyre right. FYO the mass of the sun is 1.989 x 10^30
 
  • #6
Yeah - you can do F is proportional 1/r^2 and then compare it with the centripetal force F = (mv^2) / r (m is the mass of the planet that's orbiting)
 
  • #7
futb0l said:
Yeah - you can do F is proportional 1/r^2 and then compare it with the centripetal force F = (mv^2) / r (m is the mass of the planet that's orbiting)

thats what i did, but when i compare radiuss the forcves don't match up. My reasoning is if r EARTH = 10000000 (obviosuly not right) and r MARS = 20000000, then ratio is 1:2, and the force of Mars will be 1/4 ( 1/(2r)^2 )the force of earth, providing F proportional to 1/r^2 is correct
 
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  • #8
The masses of the planets are not an issue in this problem. The centripetal acceleration you started with is just that, an acceleration. The centripetal force is the centripetal acceleration times the mass of the planet. Equate that to the gravitational force, and the mass of the planet divides out.
 
  • #9
i don't follow.
"Equate that to the gravitational force, and the mass of the planet divides out."
what does this mean? if my calculations are right, then why don't my forces match up with the inverse square law? is there a step I am missing or something?
 
  • #10
Assume the inverse square law is valid. The gravitational force

[tex] F = \frac{Gm_Sm_P}{r^2} [/tex]

is the the centripetal force acting on the planet

[tex] F = \frac{m_Pv^2}{r} [/tex]

These must be equal

[tex] \frac{Gm_Sm_P}{r^2} = \frac{m_Pv^2}{r} [/tex]

Divide both sides by the mass of the planet

[tex] \frac{Gm_S}{r^2} = \frac{v^2}{r} [/tex]

Multiply both sides by r squared

[tex] Gm_S = v^2r [/tex]

Take the square root

[tex] \sqrt{Gm_S} = v \sqrt{r} [/tex]

So, if the inverse square law holds, the planetary velocity times the square root of the distance from the sun is a constant. Planetary mass divides out of the calculation.

A table of distances and velocities is available here. Check it out.

http://www.spirasolaris.ca/sbb4b.html
 
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  • #11
Matheson said

"then you can compare the radius's and come up with a ratio of about 1:1.87 for radius's of Mercury:Venus"
"so we should then be able to say that (using F proportional to 1/r^2) the force of Venus should be 1/(1.87^2) the force of Mercury, correct?"

NO. The force of gravity on a planet is proportional to its mass. This is where you went wrong I think. You cannot take the ratio of the radii of the orbits alone to find the force of a second planet from the force on the first planet. What you did was to calculate the force that would have been on one planet if you had moved it to another planet's orbit.
 
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  • #12
Whats to be done here is to use

mv^2/r =GmM/r^2

Because it's the centripetal force which resists the attraction towards the sun which depends on the mass of the orbiter and the sun.
 
  • #13
kusal said:
Whats to be done here is to use

mv^2/r =GmM/r^2

Because it's the centripetal force which resists the attraction towards the sun which depends on the mass of the orbiter and the sun.

The consequences of the equality in your post is what is all worked out two messages back. Centripetal force does not resist the attraction towards the sun. It is the attraction toward the sun necessary to hold the planet in orbit.
 
  • #14
thanks HEAPS to older dan for that little proof, that makes complete sense! i think the only problem is the question is set out like "find this and this and this and hence prove the inverse square law"
im pretty sure the teacher wants me to work out some forces, and then compare them to other forces, but as some have stated it won't be proportional because force relies on mass, which is not constant.

Thanks a ton to all who have helped, if anyone still has anything to add it would be greatly appreciated! I am still a little fuzzy on how I am suppost to do it with the hence part, would it help if i posted the exact wording of the question?

thanks again,
-matt
 
  • #15
matheson said:
I am pretty sure the teacher wants me to work out some forces, and then compare them to other forces, but as some have stated it won't be proportional because force relies on mass, which is not constant.

I suspect your teacher would be happy to see you work out the implications of the inverse square law algebraically to reach a conclusion that orbits all have something in commom. The way I did it resulted in a product of a variable and a square root of a variable being constant. Let me suggest that you use the same starting equations to calculate the period of the orbit for each planet and show that the data agrees with the calculations.
 
  • #16
so that's using T= 2(pi)r/v right?
and is that sut to show r and v are correct?

and does anyone else have any other ideas about doing the queisotn with the 'hence' sort of approach?
 
  • #17
on a bit on an unrelated topic, how do you post those equations and terms and stuff in the larger black text?

but back on topic, is the above asumption, sut = used, sorry for the bad spelling. anyway anyone still avaiable to help me out?
 
  • #18
The equations are done with LaTex. This is my first time using this stuff. You can get info here

https://www.physicsforums.com/showthread.php?t=8997&highlight=LaTex

I didn't mean to ignore your previous note. I'm still learning my way around here and didn't find my way to page 2 of the dialog until now.

As for the periods of the planets, you should be able to predict the relationship between period and distance from the sun by eliminating v from the equations. The average v is, after all, the distance the planet travels in its orbit per unit time. The circumference can be easily expressed in terms of radius, and data for the periods is readily available for verification. Basically, what I am suggesing you do is to come up with Kepler's third law by approximating the orbits as circles.

http://home.cvc.org/science/kepler.htm
 
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  • #19
so how would that help prove the inverse square law?
is T^2 (propotional to) r^3 related to F (proportional to) 1/r^2 ? sorry for all the questions I am just really not catching on...

also, for T^2 (propotional to) r^3, is r the same value used for the radius of a planets orbit? i noticed on the site you gave me the terminology "mean distance (semi-major axis) from Sun" .
and when i tried comparing a 'r' of Earth in metres ( 149600000000 ) to the period is seconds ( aprox 31493080 with rounding calcualtions) they are definatelly not equal, or anywhere close to it. i feel stupid for asking but what am i doing wrong here?
 
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  • #20
If you look back to my derivation of [itex] \sqrt{Gm_S} = v \sqrt{r} [/itex], the starting point for that calculation was the ASSUMPTION that the inverse square law was valid. I could have made the problem more abstract by assuming I did not know the constants in the equation. The important thing is that the conclusion was based on the assumption of an inverse square force. Any other assumption would have led to a different conclusion.

Other equivalent conclusions can be reached from the same starting point, or by modifying my result. The math is not easy for elongated eliptic orbits, but it turns out that many of the conclusions reached about circular orbits hold true for eliptic orbits as well. We are not trying to prove that here. All we are doing is showing that for circular orbits the inverse square assumption leads to certain conclusions. Since the planetary orbits are nearly circular, we can approximate them as circles.

If you square both sides of my result you get [itex] Gm_S = v^2 r [/itex]. If you compute v in terms of the circumference of the orbit and and its period you get [itex] v = 2 \pi r / T[/itex], so [itex] Gm_S = 4 \pi^2 r^3 / T^2 [/itex] or [itex] T^2 = [\Left 4 \pi^2 / Gm_S \Right] r^3[/itex]. I'm not suggesting you go through my result to get to the last equation. Go back to equating the inverse square gravitational force with the centripetal force, replace the v with circumference divided by period, and take the shortest path you can to show that it leads to the conclusion that the square of the period is proportional to the cube of r. If you had not started with the assumption of inverse square force, you would not have gotten to "the period squared is proportional to radius cubed" conclusion. If the data agrees with the conclusion, then the assumption is justified.

You don't have to use the result to calculate the period of the orbit of any planet from its distance from the sun. If you know all the constants, you can do it, but suppose you did not know the constants and the only data you had was the periods of the planets and their distances from the sun. The last equation above implies that the ratio of the period squared to the distance cubed is a constant. Look up the data for the nine planets and calculate that ratio for each of them. Don't try to calculate the period from the distance, or vice versa; just calculate that ratio. If that ratio is very nearly the same for all the planets, then you have confirmed your conclusion, and since the conclusion was based on the assumption of inverse square, your data has supported the theory.

It is OK to use the mean distance from the sun, or the semi-major axis for r. The orbits are nearly circular, so those numbers are good approximations for the distance from the sun.
 
  • #21
i underdstand you putting ALOT of effort into explaining this to me, but i still don't really get it.
Why can't you compare a planets period in seconds to its radius is metres and come up with a T^2 r^3 ratio? like it doesn't work.

why do we try to prove keplers third law? how will that mean F (proportional to) 1/r^2 ?
i understood up to the first equation, but to my understanding wouldn't that have proven F (proportional to) 1/r^2 already? As in if that's correct, then F MUST be proportional to 1/r^2. (im talking about the square root one with Gm = vr with square roots in there).

the main question is just why do you have to show T^2 is proportional to r^3
 
  • #22
Showing that T^2 is proportional to r^3 proves keplers third law.
"The square of a planet's orbital period is proportional to the cube of the semimajor axis of its orbit"

In order to keep an object in circuluar path, then a force of:

[tex] F = \frac{mv^2}{r} [/tex] (1)

is required to keep the object at a velocity (v) at path of radius (r).

And as you know, the gravitational force is given as:

[tex] F = \frac{Gm_1m_2}{r^2} [/tex] (2)


So then in order for an object to stay in a circuluar path both of those forces must equal each other. So plugging (1) into (2) you have:

[tex]\frac{mv^2}{r} = \frac{Gm_1m_2}{r^2}[/tex] (3)

Now if you solve (3) for v you get:

[tex] v^2 = \frac{Gm_1}{r} [/tex] (4)

Now the time to complete one orbit is given as:

[tex] T = \frac{2\pi r}{v}[/tex]

If you solve this for v you get the following:

[tex] v = \frac{2 \pi r}{T}[/tex]

Now if you plug this into (4) look what you get:

[tex] \left(\frac{2 \pi r}{T}\right)^2 = \frac{Gm_1}{r} [/tex]

Solve this for T and low and behold:

[tex] T^2 = \frac{4 \pi^2 r^3}{GM}[/tex]
 
  • #23
matheson said:
i underdstand you putting ALOT of effort into explaining this to me, but i still don't really get it.
Why can't you compare a planets period in seconds to its radius is metres and come up with a T^2 r^3 ratio? like it doesn't work.

why do we try to prove keplers third law? how will that mean F (proportional to) 1/r^2 ?
i understood up to the first equation, but to my understanding wouldn't that have proven F (proportional to) 1/r^2 already? As in if that's correct, then F MUST be proportional to 1/r^2. (im talking about the square root one with Gm = vr with square roots in there).

the main question is just why do you have to show T^2 is proportional to r^3

You don't HAVE to show that T^2 is proportional to r^3. It is something that falls out by doing algebraic manipulations on the starting assumption that circular planetary orbits are the result of a force that is an inverse square law force. From this point of view, it is a prediction based on a theory, the theory being inverse square law. To verify a theory, you need to see if the data supports the predictions based on that theory. You have the data for nine planets. You know their periods and you know to a good approximation the radii of their nearly circular orbits. The theory predicts that T^2/r^3 should be the same for every planet. Calculate the ratio for each planet and compare them. Does the data support the theory?

Perhaps it would help you understand if you developed a counter theory or two. What would be the predicted relationship for T^n/r^m if you assumed gravity was inversely proportional to r^3 or to r instead of r^2. Do the measured values of T and r support these counter theories?

I don't know why you are not getting the right values for T based on r using the known constants. If you post the details of your calculation for one of the planets, perhaps we can see where things are going wrong.
 
  • #24
with T^2 proportional to r^3, how do you put values in?
is it

(T^2) = (r^3)

or

(T^2) / (r^3) = same for all planets?

because when you sub in values for Earth with the first one, it doesn't equate.
T = 31 493 080s
r = 149 600 000 000m

already you can see T^2 won't = r^3
am i using the wrong units or am i using keplers third law incorrectly?
 
  • #25
matheson said:
with T^2 proportional to r^3, how do you put values in?
is it

(T^2) = (r^3)

or

(T^2) / (r^3) = same for all planets?

because when you sub in values for Earth with the first one, it doesn't equate.
T = 31 493 080s
r = 149 600 000 000m

already you can see T^2 won't = r^3
am i using the wrong units or am i using keplers third law incorrectly?

Saying that two quantities are proportional is not saying they are equal. You certainly cannot expect time squared to be equal to distance cubed. Time and distance are not the same kind of thing. When two quantities are proportinal, it means their ratio is always the same, or, equivalently, that one is a constant times the other. The equation that comes from the inverse square gravity and centripetal force connection is

[tex] T^2 = [\Left 4 \pi^2 / Gm_S \Right] r^3[/tex]

If you put the radius of some planet's orbit cubed into the right hand side, and multiply by the constants in [] you should get that planet's period squared. But even if you did not know those quantities in [] you could still know that they combine to some constant value. Then you could use the period and radius of one planet to figure out what that constant is. And then you could use the value of that constant to find the period of any other planet if you knew its radius. Alternatively, just knowing (or believing) that T^2 and r^3 are proportional means that the ratio of those two things is (or is predicted to be) a constant, so if you calculate the ratio for each planet you should get the same result every time.
 
  • #26
matheson: Post your entire calculation process. Not just the answer. Post why you think something is a certain way, ie: why did you think T^2 is proportional to R^3? ... who cares if you are wrong, that is fine. I definitely don't know what I'm doing half the time, and I'm sure a good portion of the people on this forum are the same way. Part of learning is making mistakes. But making the same mistake over and over again is not helpful to yourself. A big part of physics (from the little I know, and have had) is having a good approach to fixing your mistakes.

So, here's an example of a possible though process to an equation. (Please bear with me here, this might be a little tedious and not needed, but it will be easy to understand and therefore the "heart" of this example will not be lost in interpretation.

ok... so we have:
T^2 = R^3
hmmm... time^2.
Well time is something that occurs, at a steady pace (not getting into reference frames, near speed of light calculations, yadda yadda)... and it squared means that the quantity of itself is growing very fast. Like if we assume that T is measured in seconds, and we plug in 5 seconds into T^2 we have 25 seconds. If we plug 10 seconds into T^2 we have 100 seconds... cool... ok that's what T^2 does... it grows fast.

Now R^3... the radius to the power of 3. So we plug in a small radius and boom, a huge radius is returned... ok... so that's what R^3 does... it grows even faster then time.

ok so now if we just analyze time, then we have (i.e. getting rid of any mathematical operator on our variable that we are interpretting.)

T=sqrt(R^3)... so let's say we plug in a radius of 5... well we know that R^3 grows really fast... and of course we can compute it as 125 that's simple. The thing is though. Yes in this case it is quite simple to compute 5 to the power of 3, but a lot of times we will not have numbers... We will just have to understand what the variable means. Most times it is best anyhow, not to use "numbers" until the end, so that means you HAVE to understand each expression, so when you do the robotic chore of plugging in numbers, you understand WHY you are plugging those numbers in.

so anyhow...
Now we just have the sqrt() to deal with. So the sqrt is asking for a number multiplied by a number equals the square rooted number (in this case 125), so hmm I don't have a calculator... but its around 11.2... so yeah... we plug in a radius of 5 and out comes roughly 11.2.

yes, ok... so plugging in numbers yielded.

11.2 seconds (or whatever time unit) = 5 meters (or whatever radius)

?

now that doesn't make any sense.. you can't say red = blue.

ok so yeah, I killed a really simple eqauation with a lot of detail, but my point is this... when you begin to think about what everything is doing to one another, you can start to narrow down your "problem" into something formidable. So give that a shot for your expressions and see what you are coming up with. Like, many times you mentioned the 1/r^2 "law" well what does that mean? ... what does that mean when you are comparing it to objects that uphold to this same "law"...

so anyways... write down your expressions, and answers for everything.. understand each expression, and narrow down your problem...yadda yadda
 
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  • #27
hey thanks guys, I've been away for a while.
im pretty sure I am understanding it all now, so thanks heaps for all your help. ill see if i can give my teacher like a draft, to make sure what I've done is what he wanted, and if it isn't ill try to figure it out, but if i can't expect to see me back here!
 

Related to Understanding the Inverse Square Law in Planetary Motion

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