# Need help with Gravitational Constant (G)

1. Apr 25, 2012

### eggman100

With this formula, you should apparently be able to work out the schwarzschild radius with 2GM over c^2

but it shows the gravitational constant on google being:

G = 6.67300 × 10-11 m3 kg-1 s-2

Can someone please explain this to me? - Because then I would be able to work out the schwarzschild radius and thus then I would be able to work out the gradient of a graph showing at distance x and x+Δx I would be able to show in percentage the decrease of the effect of the gravitational field of the mass (being a planet) on space and time in terms of an object exiting the surface of the planet into space until it is no longer affected (that is my plan) but i need someone to explain how the m^3 kg^-1 s^-2 works? - How do i represent that as my final answer????? and can the index number (+ decimal) be different? as in, is that number relative to earth??? or the actual number? - I think its relative to a group that actually researched the number to x number of decimal places :\$ -

Please help with the notation, thank you ever so much readers and physics guys :D much respect :)

2. Apr 25, 2012

### Nabeshin

This is just the units that they've quoted the gravitational constant in. In particular, these are SI units where lengths are in meters, masses in kilograms, and times in seconds. If you use these units, the formula for schwarzschild radius gives an answer back in meters.

3. Apr 25, 2012

### tiny-tim

hi eggman100!

G comes from Newton's law of gravitation …

acceleration = GM/r2,​

where M is the mass (of the star, etc) and R is the distance

acceleration has dimensions of distance/time2 (L/T2),

so the units (of measurement) of that equation are

L/T2 = G times M/L2

so G must have dimensions of L3/MT2,

so in SI units, G is measured in m3/kg.s2

(and GM/c2 has dimensions of (L3/MT2)M/(L/T)2, = L = distance)

4. Apr 25, 2012

### eggman100

Thank you, but i still dont understand, so how would i use this to work out the schwarzschild radius of the earth? I know its around 9millimeters, but so what it is:

2xGx(6x10^24)
-------------- (mass being rounded and c^2 being rounded)
300,000^2

What would i put in place of G? - and so you mean that in terms of the schwarzschild radius, what does it mean? - so your saying that with an increase of one meter from the mass (surface of planet) then your saying, per one second, per one meter, there is a change of X kilograms? and going over a scale of time (seconds) ??

5. Apr 25, 2012

### eggman100

and i understand the m^3- but how would you change your answer according to kg x s^2? - Can someone please give me an example I can understand? something like earth please? but I still thank you guys for your help :D - but a little more help would be much more appreciated :D

6. Apr 25, 2012

### tiny-tim

hi eggman100!

(try using the X2 button just above the Reply box )
you have to substitute 6.67300 × 10-11 for G in that equation,

and that gives you the schwarzschild radius, R, in metres

if the mass of the earth was twice as large, then R would be twice as large

(btw, in Newton's law, the distance is measured from the centre of the body, not from the surface)

if the earth somehow shrunk (keeping the same mass) to less than R, then it would be a black hole

7. Apr 25, 2012

### Waterfox

Don't let the units confuse you, in place of G just use 6.67*10^-11
Also your value for c^2 is wrong, it should be 300,000,000^2 = (3*10^8)^2 = 9*10^16

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