Exploring G: The Gravitational Constant

In summary, G is the gravitational constant. It is approximated to be 6.67428pm0.0010*10(power-11). It has units of m*kg*s.
  • #1
questioner
10
0
G is the gravitational constant. it's approximate value is:
G = (6.67428[tex]\pm[/tex]0.0010)*10(power -11) m(power 3) kg(power-1) s(power-2)
= (6.67428[tex]\pm[/tex]0.0010)*10(power -11) N m(power 2) kg(power-2)
= (6.67428[tex]\pm[/tex]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 don't know how to type the power sign so i put (power 3) etc everywhere. sorry.
 
Physics news on Phys.org
  • #2
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
because it's value isn't just a plain number. i don't know what all those signs do there:[tex]\pm[/tex], 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
http://www.google.com/search?q=G

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 [itex]\pm [/itex] 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:
  • #5
so the [tex]\pm[/tex] sign means that you don't 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
questioner said:
so the [tex]\pm[/tex] sign means that you don't 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?
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?

This may be helpful:
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.)

and twhat are the m kg and s for?

http://hyperphysics.phy-astr.gsu.edu/hbase/units.html
 
Last edited by a moderator:
  • #7
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
oh...
ok.
but that means that its value is aproximately 6.6e-11?
 
  • #9
questioner said:
oh...
ok.
but that means that its value is aproximately 6.6e-11?

Almost, it's approx:

[tex]G = \frac{6.67*10^{-11}m^{3}}{kg*s^{2}}[/tex]

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

1. What is the gravitational constant?

The gravitational constant, denoted by the letter G, is a fundamental constant in physics that represents the strength of the gravitational force between two objects. It is an important value in understanding the behavior of objects in the universe and is used in many equations in the field of physics.

2. How is the gravitational constant determined?

The gravitational constant is typically determined through experimental methods, such as measuring the gravitational force between two objects of known masses and distances. This value can also be calculated using other physical constants, such as the mass and radius of the Earth, or by using data from astronomical observations.

3. Why is the gravitational constant important?

The gravitational constant is important because it helps us understand the behavior of objects in the universe. It is used in the calculation of the force of gravity between objects, as well as in the study of celestial bodies and their motion. It is also a key factor in the theory of general relativity, which explains the relationship between gravity and the geometry of space and time.

4. Has the value of the gravitational constant always been the same?

No, the value of the gravitational constant has not always been the same. It was originally thought to be a universal constant, but as technology and measurement techniques have advanced, more precise values have been determined. The currently accepted value of G is 6.67430(15) × 10^-11 m^3 kg^-1 s^-2.

5. How does the value of the gravitational constant affect the behavior of objects in space?

The value of the gravitational constant determines the strength of the gravitational force between two objects. This force is responsible for the motion of objects in space, such as planets orbiting around a star or moons orbiting around a planet. A higher value of G would result in a stronger gravitational force, while a lower value would result in a weaker force.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
665
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
507
  • Introductory Physics Homework Help
Replies
3
Views
937
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
250
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
2
Replies
63
Views
2K
Replies
8
Views
763
Back
Top