## year 12 physics assignment

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?

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 Recognitions: Homework Help Science Advisor 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.
 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!

## year 12 physics assignment

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:
$$G = 6.67\times 10^{-11}Nm^2/kg^2$$

 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
 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)

 Quote by 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)
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

 Recognitions: Homework Help Science Advisor 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.
 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?
 Recognitions: Homework Help Science Advisor 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
 Recognitions: Homework Help Science Advisor 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.
 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.

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 Quote by 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.
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.

 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

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