The Equation for Classical Gravity - Am I using this right?

[metacogitans]

 

Where G is the Gravitational Constant 6.674×10⁻¹¹ N⋅m²/kg²

m1 and m2 are the masses of the two objects in kilograms

and r^2 is the distance between them squared

 

I tried a test problem using the Earth and the Moon, and had some questions.

My values were:

moon:7.34767309 × 10^22 kilograms
Earth:5.972 × 10^24 kg

distance between their centers of mass: 384,400,000 meters

 

I ended up getting this for an answer using the calculator built into the Google search engine: 2.9285715e+37 which didn't look right, so I did it again but to do it faster I only used a few of the decimal places in the numbers, and got 7.4778663e+28

 

I've had problems with this calculator before when trying to get it to use scientific notation; I did use parantheses correctly where appropriate and correct order of operations; it's just a poorly coded calculator. Can anyone recommend me an online calculator or calculator software that works well, and will display many digits of decimals? Being able to use the symbols / * ^ ( ) and able to have many sets of parantheses inside each other. All of that is absolutely Essential for a calculator to avoid having to enter the equation one piece at a time. I have a Texas Instruments Graphing Calculator but it's out of batteries I've wanted calculator software on my computers for a long time though, might as well find a decent one now.

 

Anyways, a few other questions:

 

If the number 7.4778663e+28 is close to correct for the gravitational force between the Earth and the Moon, I'm guessing that I have to divide it by the kilograms then of whichever celestial object to get a value in meters per second? What do I do exactly with "N * m^2 / kg^2"?

 

is r^2 in the equation really the distance between centers of mass squared? Why is the symbol r being used? A few sites said it was something different (one said it was the distance from the surface of the objects, and few pictures on google image search said 'radius of the moon' (which would almost be correct for the moon's surface, but is still wrong).

Edited November 28, 2015 by metacogitans

[Strange]

Wolfram Alpha is pretty good for this sort of thing. It (like Google) also doubles up a search engine so you can enter things like:

"G * (mass of moon) * (mass of earth) / (distance earth to moon in km)^2"

http://www.wolframalpha.com/input/?i=G+*+%28mass+of+moon%29+*+%28mass+of+earth%29+%2F+%28distance+earth+to+moon+in+km%29^2

 

This gives an answer of 2x10²⁰ N.

 

 

If the number 7.4778663e+28 is close to correct for the gravitational force between the Earth and the Moon, I'm guessing that I have to divide it by the kilograms then of whichever celestial object to get a value in meters per second?

 

Dividing the force by mass will give you acceleration.(*) To get a speed from that, you need to know how long the acceleration is applied for. Or are you trying to calculate orbital speed? https://en.wikipedia.org/wiki/Orbital_speed

 

(*) The fact that you divide by mass (which therefore cancels out) is the reason that all objects fall at the same rate (as shown by Galileo).

 

 

is r^2 in the equation really the distance between centers of mass?

 

Yes. If something says it is 'radius of the moon' then it is probably calculating the force of gravity at the surface.

 

Why r? Perhaps because it was originally used to work out orbits?

If you want to know how to derive orbital speed, this might be useful: http://math.stackexchange.com/questions/255694/deriving-an-equation-for-the-orbital-period-of-a-satellite

[metacogitans]

Here we go, I found one of the good ones, heh check it out:

I got this for an answer now (I cut off some of the decimals, didn't need 5 dozen of them): 198,193,345,664,504,055,673.88288950657

 

 

Here's the link to that calculator if anyone else wants to check out: http://web2.0calc.com/

I really like how it professionally organizes your equations automatically up above the calculator; and your answer can be copied to your clipboard out of the entry bar (copy/paste is a HUGE bonus imo).

Wolfram Alpha is pretty good for this sort of thing. It (like Google) also doubles up a search engine so you can enter things like:

"G * (mass of moon) * (mass of earth) / (distance earth to moon in km)^2"

http://www.wolframalpha.com/input/?i=G+*+%28mass+of+moon%29+*+%28mass+of+earth%29+%2F+%28distance+earth+to+moon+in+km%29^2

 

This gives an answer of 2x10²⁰ N.

 

 

Dividing the force by mass will give you acceleration.(*) To get a speed from that, you need to know how long the acceleration is applied for. Or are you trying to calculate orbital speed? https://en.wikipedia.org/wiki/Orbital_speed

 

(*) The fact that you divide by mass (which therefore cancels out) is the reason that all objects fall at the same rate (as shown by Galileo).

 

 

Yes. If something says it is 'radius of the moon' then it is probably calculating the force of gravity at the surface.

 

Why r? Perhaps because it was originally used to work out orbits?

If you want to know how to derive orbital speed, this might be useful: http://math.stackexchange.com/questions/255694/deriving-an-equation-for-the-orbital-period-of-a-satellite

Thank you very much, that clarified everything I needed to know

I was looking for acceleration; I forgot to say meters per second "squared".

I'll check out orbital speed next, thanks for the link

Edited November 28, 2015 by metacogitans

[imatfaal]

It is good form to group top and bottom together ie GMm then over r^2 . Having another part of the top after the bottom risks a missing bracket and the last multiple being put on the bottom.

 

For sums like this it is often easier to separate the powers of ten and the coefficients - ie

 

(22+24) - (8*2 + 11) = 46 - 27 = 19

 

6.674*5.972*7.374/(3.844^2) = 19.89

 

so the answer is 19.89 x 10^19 or 1.989 x 10^20 N

 

this means you can do the first bit just handling the powers of ten and get a very reasonable estimate without using anything but your mental arithmetic