1. May 31, 2008

### Razza

:uhh:

1. The problem statement, all variables and given/known data

I'm trying to figure out the mass of the earth.

2. Relevant equations
I know what it actually is, 5.97*10^24 kg, but it's just how to do it that's got me stumped.

I know the F = Gm1m2/d squared (or r squared depending what you're taught) formula.

3. The attempt at a solution

I have the mass of my object, which is 0.012kg, i know my d, that's 6.3781*10^3 squared (which is 4.06801586*10^7), I know G (6.67*10^-11), but I'm just confused about the F, in my book we worked it was 931 N = ...., but I think that was just 9.8 N (gravitational acceleration) multiplied by my teacher's weight, as an example.

If I do 0.012*9.8 I get 0.1176 N, so does that mean my equation should look like:

0.1176 N = (6.67*10^-11) * 0.012 * Me (mass of earth)
6.3781*10^3​

Then I rearrange the formula to get
0.1176*(4.06801586*10^7)
(6.67*10^-11)*0.012

But the answer I keep getting, no matter how many different ways I try, even when I do get the right answer, I'm about 1 000 000 off, i.e. I get 5.97*10^18. WHY oh why is this happening? I have to show working for my assignment. Now I'm lost when I even follow my own book cuz we used different masses i.e. 60 kg instead of 12 GRAMS, (the 12 grams is the weight of my sinker).

2. May 31, 2008

### Cvan

Units are your downfall. You're really close. Always watch units.

(Think..what could be off to give you a factor of 10^6 off in terms of order of magnitude?)

3. May 31, 2008

### konthelion

Given the radius of the Earth and G constant $$G\ =\ 6.673(10)\ \times\ 10^{-11}\ m^{3} kg^{-1} s^{-2}$$, then by the second formula
You get
$$F=\frac{Gm_{earth}m_{1}}{R^2}=m_{1}g$$, where g is acceleration due to gravity

Now, divide both side by $$m_{1}$$ to cancel out the mass m1. We don't need it.
Then, we get $$m=gR_{earth}^2/G$$. Plug and chug.

Last edited: May 31, 2008
4. May 31, 2008

### Razza

Umm...

That I haven't converted something properly? I thought about how I converted g to kg, but that all seems right...

Maybe I have the radius of the earth wrong.. is it meant to ... nah its not meant to be in meters is it?

(Thank you by the way )

5. May 31, 2008

### Cvan

Check the units of the newton (kg*m)/sec^2
and the units of G : m^3 kg^-1 s^-2

and you will have learned your first (and hopefully forever memorable) lesson that you should ALWAYS include units.

6. May 31, 2008

### Razza

Konthelion, yes that works! But... I'm still getting to the power of 18, instead of 24. I'm still that million off, and I cannot figure what I'm doing wrong...

(Thank you )

7. May 31, 2008

### konthelion

Check your distance d i.e. radius of the Earth, it is $$6.37 × 10^6$$ not $$6.37x10^3$$

8. May 31, 2008

### Razza

Seconds squared is acceleration right? Um..

Wait..
Conversion.. Umm... 1000 squared is one million.. so... umm, yeah the

WAIT I think I've got it...

I converted the radius from km to metres.... and it gave me 4.068015*10^13, instead of 10*7, which is the 6 more decimals I need..

Oooh thank you thank you thank you I'm going to see if it works now, it should!!:rofl:

9. May 31, 2008

### Razza

Ooh, So does that mean...
Yes thats it... I'm sure...

Thank you, both of you. Soo much

I just have to find my paper with it written haha I have so much working out here...

Let's see if I do this right...

10. May 31, 2008

### Razza

YESSS

It worked!!

Aww thank you enormously! THANK YOU! YAY!

I know it seems ridiculously small that I shouldn't get something like this, but thank you.

Thats all it was, the units! Thank you so much!