astrophysicist Help?

[power5000]

hey from what I\'ve seen there are 2 kinds of ksp players people who build crazy wacky rockets and do it just for fun then the more RP style players who like doing the math and figuring out that type of stuff turns out I\'m the later but being in highschool I don\'t know any of that math really.... so I was thinking why not post on the forums and just have a common post of knowledge for people to ask questions and maybe so awesome smart people to help us answer them?

Question: 1 F=G(m1*m2/r^2)

first off I\'ve been reading the page about basic astrophysics and it says '

The third law states that if body 1 exerts a force on body 2, then body 2 will exert a force of equal strength, but opposite in direction, on body 1. This law is commonly stated, 'for every action there is an equal and opposite reaction'.

In his law of universal gravitation, Newton states that two particles having masses m1 and m2 and separated by a distance r are attracted to each other with equal and opposite forces directed along the line joining the particles. The common magnitude F of the two forces is

F=G(m1*m2/r^2)

where G is an universal constant, called the constant of gravitation, and has the value 6.67259x10-11 N-m2/kg2 (3.4389x10-8 lb-ft2/slug2).

Let\'s now look at the force that the Earth exerts on an object. If the object has a mass m, and the Earth has mass M, and the object\'s distance from the center of the Earth is r, then the force that the Earth exerts on the object is GmM /r2 . If we drop the object, the Earth\'s gravity will cause it to accelerate toward the center of the Earth. By Newton\'s second law (F = ma), this acceleration g must equal (GmM /r2)/m. g=GM/r^2'

now I understand everything up until 'If we drop the object, the Earth\'s gravity 'ect... is g=GM/r^2 the same as F=G(m1*m2/r^2) if it\'s not or im missing something basic i\'m sorry -.-...

Answered from post 2 through 12

[RevJB]

Hey Power,

I read physics at university/college so hopefully i can help.

You asked...'is g=GM/r^2 the same as F=G(m1*m2/r^2)?\'

They are similar, and this can trip people up but they are different in a subtle way. g in this case is the acceleration the object is experiencing and F is the Force. They are related by the equation F=ma which it looks like you have already come in contact with. So you can see that instead you could write a=GM/r^2 instead. Essentially we calling the acceleration towards the Earth g.

I hope that helps, if i have the wrong end of the stick and it\'s something else that you are unsure about let me know, i\'d be glad to help.

RevJB

[Cykyrios]

Hey,

You have F=G(m1*m2)/r² and F=ma, where a is g, so you can write G(m1*m2)/r²=mg, hence g=G(m1*m2)/mr². Since m1=m and m2=M (or the other way around), g=GmM/mr²=GM/r².

You can\'t really say that 'g=GM/r^2 is the same as F=G(m1*m2/r^2)' but rather that the former derives from the latter in the case of Earth\'s gravity, and as RevJB said, the force and the acceleration are related by the mass.

[BlazingAngel665]

I am a high schooler too, however, a trip to wikipedia, a calculator and MechJeb make all the diffrence, and MechJeb taught me really useful things like which ways to point you rocket so now I don\'t need it (all the time . . . . )

[power5000]

Thank you everyone for your help I understand it alot more now but what are each one used for (an example for each if possible (other then f=ma)

P.S. if anyone has question please ask them I\'ll add them to the op

[power5000]

bump?

[Bluejayek]

We use g for simple math near the surface of the planet mostly, since the altitude is nearly constant. g=9.8m/s/s at the surface of the earth, and you an use this formula for things like projectile motion. You can also talk about the value of g at different altitudes. The same value also aproximately works in LEO, as an altitude of say 100km compared with the earths 6000km radius is only a correction of 4% or so.

Basically the formula F=mg where g=9.8 is a quick aproximation, but the more accurate is always to use F = m1m2G/r^2.

Using the general formula g = mG/r^2 for acceleration is convenient as it shows that all bodies in orbit around the same object will have the same orbits, as it is acceleration whihc determines the orbit, not force directly.

[power5000]

okay so F=G(m1*m2/r^2) lets say we have a ship with a mass of X or M1 we have kerbins mass of 5.29*10^22 or M2 and / r^2(our orbit+kebins radius=700,000) and G=6.67*10^-11 witch would

so X=10

(6.67e-11)*(10*(5.29e22)/700000^2)=72.0087755102040816

correct?

[Bluejayek]

Yup, simple substitution.

[power5000]

Okay thank you very much it seems all so stupid that i didnt understand that now -.- one question what is 72.0087755102040816? is it the m/s or something that kerbin will pull my ship back?

[Bluejayek]

Its the force at that exact distance. What is really more meaningful is the acceleration (a = F/m where m is your ships mass). Then, you can compare it with the centripetal acceleration (a_c = v^2/r) to figure out the velocity needed to maintain a stable orbit.

v = sqrt(r*F/m)

This is a stable circular orbit by the way.

[Kosmo-not]

I didn\'t thoroughly read through all the posts, so forgive me if someone mentioned this already.

g=GM/r^2 is a good approximation of F=G(m1*m2/r^2) when one mass is much smaller than the other (with M being the mass of the larger). This approximation is accurate for a spacecraft going around a planet. The other equation would be used in cases such as forces between a planet and its moon.

[power5000]

Its the force at that exact distance. What is really more meaningful is the acceleration (a = F/m where m is your ships mass). Then, you can compare it with the centripetal acceleration (a_c = v^2/r) to figure out the velocity needed to maintain a stable orbit.

v = sqrt(r*F/m)

This is a stable circular orbit by the way.

Thank you very much for eplaining this and @Kosmo-not yeah someone did explain that but you put it in a MUCH easier to understand fashion and thank you if anyone else has questions please ask i\'ll add them to the op

P.S.... so in (a_c = v^2/r) a is acceleration? v is velocity and r is radius but what does (a_c) or (_c) mean??

[Kosmo-not]

P.S.... so in (a_c = v^2/r) a is acceleration? v is velocity and r is radius but what does (a_c) or (_c) mean??

Centripetal acceleration, I would assume.

[Samurai20039]

Where can i find the basic constants in this universe where i can use my formulas... Such as the GM of the celestial bodies and the other universal constants such as constant of gravitation, acceleration of gravity and well you know all that orbital mechanics stuff.

[TheKerbinator]

Wikipedia has alot of information on each celestial body.

[bsalis]

Where can i find the basic constants in this universe where i can use my formulas... Such as the GM of the celestial bodies and the other universal constants such as constant of gravitation, acceleration of gravity and well you know all that orbital mechanics stuff.

Should be on the Wiki. Note that the physics are pretty much the same as RL. However in the Kerbal universe, density is much higher, such that Kerbin has the same gravity as Earth (9.8 m/s), but is very small.

[UmbralRaptor]

Where can i find the basic constants in this universe where i can use my formulas... Such as the GM of the celestial bodies and the other universal constants such as constant of gravitation, acceleration of gravity and well you know all that orbital mechanics stuff.

Since the current KSP wiki seems to be hard to find: http://kspwiki.nexisonline.net/wiki/Celestials

[power5000]

I have another question what is the formula to find the orbital period of a body (how long it will take my ship to orbit once around kerbin or the mun around kerbin ect..)

[Kosmo-not]

I have another question what is the formula to find the orbital period of a body (how long it will take my ship to orbit once around kerbin or the mun around kerbin ect..)

http://en.wikipedia.org/wiki/Orbital_period

Go down to the calculation section.