# Year 12 physics assignment

1. Apr 18, 2005

### matheson

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 doesnt seem right.

in the end, i couldnt 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 doesnt the inverse square law match up with the calculated forces?

2. Apr 18, 2005

### OlderDan

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. Apr 18, 2005

### matheson

alright, these are done by hand so it might look a little messy:
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. Apr 18, 2005

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
$$F = \frac{Gm_1m_2}{r^2}$$
where:
$$G = 6.67\times 10^{-11}Nm^2/kg^2$$

5. Apr 18, 2005

### matheson

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 im not considering the masses however, cant it just be written as F (alpha) 1/r^2?
and so then shouldnt i be able to use the ratio system above?

please point out any errors, i cant see what ive 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. Apr 18, 2005

### futb0l

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. Apr 18, 2005

### matheson

thats what i did, but when i compare radiuss the forcves dont 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

Last edited: Apr 18, 2005
8. Apr 18, 2005

### OlderDan

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. Apr 18, 2005

### matheson

i dont 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 dont my forces match up with the inverse square law? is there a step im missing or something?

10. Apr 18, 2005

### OlderDan

Assume the inverse square law is valid. The gravitational force

$$F = \frac{Gm_Sm_P}{r^2}$$

is the the centripetal force acting on the planet

$$F = \frac{m_Pv^2}{r}$$

These must be equal

$$\frac{Gm_Sm_P}{r^2} = \frac{m_Pv^2}{r}$$

Divide both sides by the mass of the planet

$$\frac{Gm_S}{r^2} = \frac{v^2}{r}$$

Multiply both sides by r squared

$$Gm_S = v^2r$$

Take the square root

$$\sqrt{Gm_S} = v \sqrt{r}$$

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

Last edited: Apr 18, 2005
11. Apr 18, 2005

### OlderDan

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.

Last edited: Apr 18, 2005
12. Apr 18, 2005

### kusal

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. Apr 18, 2005

### OlderDan

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. Apr 19, 2005

### matheson

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 wont 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! Im still a little fuzzy on how im suppost to do it with the hence part, would it help if i posted the exact wording of the question?

thanks again,
-matt

15. Apr 19, 2005

### OlderDan

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. Apr 19, 2005

### matheson

so thats 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. Apr 20, 2005

### matheson

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. Apr 20, 2005

### OlderDan

The equations are done with LaTex. This is my first time using this stuff. You can get info here

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 [Broken]

Last edited by a moderator: May 2, 2017
19. Apr 20, 2005

### matheson

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 im 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?

Last edited: Apr 20, 2005
20. Apr 21, 2005

### OlderDan

If you look back to my derivation of $\sqrt{Gm_S} = v \sqrt{r}$, 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 $Gm_S = v^2 r$. If you compute v in terms of the circumference of the orbit and and its period you get $v = 2 \pi r / T$, so $Gm_S = 4 \pi^2 r^3 / T^2$ or $T^2 = [\Left 4 \pi^2 / Gm_S \Right] r^3$. 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.