# Some trouble with calculations…

1. May 26, 2005

### Lucretius

I was doing the Inverse Square Law, trying to find out g at the surface of the Earth.

$$\frac{4\pi(G)(M)}{4\pi(r)^2}=g$$

G = Gravitational Constant
M = Mass of Earth

$$\frac{4\pi(6.67x10^-11)(6.0x10^24)}{4\pi(6378^2)}=g$$

$$g=9,838,028$$

If I did the math right, can anyone tell me what this value means? I don't know what gravity is supposed to be quantified as. Can't be acceleration unless I did the math wrong (which I wouldn't doubt I did.) Any help is appreciated.

2. May 26, 2005

### jdavel

It looks like you're mixing units.

3. May 26, 2005

### Lucretius

Should I be getting 9.8 as my answer?

4. May 26, 2005

### juvenal

Your answer will be completely dependent upon your units. F = ma = GmM/r^2, so g = a= Gm/r^2. Acceleration has units of distance/time^2.

5. May 26, 2005

### Lucretius

So is it legal for me to use these units:

$$\frac{4\pi(6.67\times10^-11m^3 k^-1 s^-2)(6.0\times10^2^4kg)}{4\pi(6378km^2)}=g$$

If g = a, a should be m/s/s, correct?

Sorry, I'm a physics newbie :(

Last edited: May 27, 2005
6. May 26, 2005

### juvenal

Look at the units you use, and make sure they cancel out. The easiest thing to do is to convert everything to the same units. km to m, kg to g.

For example, if a quantity is 100 km/s, you would convert that to 100 km/s * 1000m/1km = 1E+05 m/s.

Do a web search on "Dimensional analysis".

7. May 27, 2005

### krab

The thing you should realize is that physical quantities are not numbers; they are numbers times units. So saying the length of your foot is 12 is simply wrong; it is 12 inches. So in the same sense, your first post made no sense. In post #5, you quote the units but do not cancel them. For example, you have a kg^-1 and a kg on top, and they completely cancel, leaving no kg in the final units. Also, the m^3 on top cancels the km^2 on the bottom, leaving metre per kilo squared. It is the kilo squared, which equals a million, that you left out. In a sense, your answer is 9838028 metres per kilo^2 per s^2, which is right, but is exactly the same as 9.838028 metres per s^2, since a kilo is a thousand (exactly).

8. May 27, 2005

### HallsofIvy

Staff Emeritus
Was there are reason for carrying that "$$4\pi$$" in both numerator and denominator?

9. May 27, 2005

### Lucretius

Okay, I think I see what I did wrong now. Let me see if I understand correctly.

$$\frac{4\pi(6.67\times10^-^1^1m^3 k^-^1 s^-^2)(7.347673\times10^2^2 kg)}{4\pi(1738km^2)}=g$$

The answer is 1.62246789 m/s/s, correct?

Last edited: May 27, 2005
10. May 27, 2005

### dextercioby

That's the number for the moon.The number looks okay,but the units do not match.Try to put everything to SI-mKs.

Daniel.

11. May 27, 2005

### juvenal

Not to mention that he's written "m/s" which is velocity not acceleration.

12. May 27, 2005

### Lucretius

On the subject of the moon, which is indeed what I was finding, I read here that this number has nothing to do with tidal effects. Do you know what would? This isn't homework by the way, it's something I am interested in finding out to show why the moon-drift argument of Young Earth Creationists is incorrect.

EDIT: I found a formula:

$$G\times M(moon)\times\frac{2R}{D^3}$$

where D = Distance from moon center to Earth center
and R = Radius of the Earth

I put all distances into meters so it would all match up.

$$(6.67\times10^-^1^1m^3kg^-^1s^-^2)(7.3\times10^2^2kg)\times\frac{2(6378000m)}{384403000m^3}=1.02\times10^-^6ms^2$$

Last edited: May 28, 2005
13. May 27, 2005

### dextercioby

How about doing those calculations again...?This time the units fit,but the numbers don't.

Daniel.