# Homework Help: What is G

1. Aug 7, 2007

### questioner

G is the gravitational constant. it's approximate value is:
G = (6.67428$$\pm$$0.0010)*10(power -11) m(power 3) kg(power-1) s(power-2)
= (6.67428$$\pm$$0.0010)*10(power -11) N m(power 2) kg(power-2)
= (6.67428$$\pm$$0.0010)*10(power -8) cm(power 3) g(power-1) s(power-2)
now what is this? is this even an absolute number? because i want to use it in the formula:

a1 = G(m2/r*r)

and i need to have an actual number for a1.

I don't know how to do it.

and i dont know how to type the power sign so i put (power 3) etc everywhere. sorry.

2. Aug 7, 2007

### Kurdt

Staff Emeritus
Its just a constant. All you'd put in your calculator when working with it is 6.67-11 or something similar. Why are you so perplexed by it?

3. Aug 7, 2007

### questioner

because it's value isn't just a plain number. i dont know what all those signs do there:$$\pm$$, m, kg, s, N and cm.
if its a constant why are they there? and how do u get 6.67(-11) out of it? aren't you ignoring the 0.001 and the other things?

4. Aug 7, 2007

### robphy

In a computer program, you might write 6.67e-11.

On a [Ti-83] calculator, you might write 6.67 [EE][(-)]11 to write 6.67E-11. [EE] is [2nd][,].
Alternatively, you might write 6.67*10^[(-)]11
or 6.67[10x][(-)]11, where [10x] is [2nd][LOG].

(Note that $\pm$ is an indication of the uncertainty associated with the measurement.
In order to consider this uncertainty and its effect on any quantity derived from it,
you have to perform an error analysis involving "propagation of errors".)

(Most physical quantities carry units, which one treats as algebraic factors.
When using a calculator, you need to be sure that you work with a consistent set of units
for the quantities you use... don't add some quantity expressed in cm with those in some other unit.
At the end of a correct calculation, you can append the appropriate units that your calculator cannot manipulate.)

Last edited: Aug 7, 2007
5. Aug 7, 2007

### questioner

so the $$\pm$$ sign means that you dont know if its 6.7 - 0.0001 or +0.0001? why don't we know that? is it because of the limits of the measurements of the physicists?
and twhat are the m kg and s for?

6. Aug 7, 2007

### robphy

Yes, you could say that.

Consider the length of a pencil on your desk.
Using some measureing device,
how would you report the length of that pencil, and your confidence in that measurement?

http://www.physics.unc.edu/~deardorf/uncertainty/UNCguide.html
http://www.upscale.toronto.edu/GeneralInterest/Harrison/ErrorAnalysis/ [Broken]
(I haven't read them in depth.)

http://hyperphysics.phy-astr.gsu.edu/hbase/units.html

Last edited by a moderator: May 3, 2017
7. Aug 7, 2007

### musicheck

Those units are there because G is used in newton's equation for gravity:
Suppose we have two objects, one with mass m1 and one with mass m2.
F= Gm1m2r-hat/r^2

where
F= the force m1 exerts on m2 or the force m2 exerts on m1 (by Newton's third law, they are equal and in opposite directions).

m1=mass of the first object
m2=mass of the second object
r=the distance between the two objects.
r-hat= a vector of length one pointing from one object to the other. When calculating the force on the object with mass m1, r-hat points from m1 to m2. When calculating the force on the object with mass m2, r-hat points from m2 to m1. This accounts for the opposite direction of the two forces.

The units of force are kg*m/s^2.
If we ignore G, we have the units kg^2/m^2 (r-hat has no units) on the right side of the equation. Thus, G must have the units m^3/(kg*s^2) to make the equation make any sense.

8. Aug 8, 2007

### questioner

oh...
ok.
but that means that its value is aproximately 6.6e-11?

9. Aug 8, 2007

### Feldoh

Almost, it's approx:

$$G = \frac{6.67*10^{-11}m^{3}}{kg*s^{2}}$$

All it does is balance out the universal law of gravitation equation, I believe.