Gravitational field strength calculation

In summary: F/m defines the strength of the gravitational field at any location, in terms of the gravitational force F that acts on a test-mass m placed at that location. (Note that we use a similar definition for electric field: E = Felectric/q.)
  • #1
Hannah7h
40
0
In what scenarios would you use the equation g= F/m instead of g=GM/r2 (or vice versa), for calculating gravitational field strength?

Update: is g=F/m used to find the force acting on a mass in a gravitational field (of strength g), whereas g=GM/r2 used to calculate the gravitational field strength at a point in the field created by the object of mass M
 
Physics news on Phys.org
  • #2
g = F/m defines the strength of the gravitational field at any location, in terms of the gravitational force F that acts on a test-mass m placed at that location. (Note that we use a similar definition for electric field: E = Felectric/q.)

g = GM/r2 is an application of that definition to the special case of the gravitational field at a distance r from a point mass, or outside a spherically symmetric mass distribution, at a distance r from the center: $$g = \frac F m = \frac {\left( \frac {GMm} {r^2} \right)} m = \frac {GM} {r^2}$$
 
  • #3
jtbell said:
g = F/m defines the strength of the gravitational field at any location, in terms of the gravitational force F that acts on a test-mass m placed at that location. (Note that we use a similar definition for electric field: E = Felectric/q.)

g = GM/r2 is an application of that definition to the special case of the gravitational field at a distance r from a point mass, or outside a spherically symmetric mass distribution, at a distance r from the center: $$g = \frac F m = \frac {\left( \frac {GMm} {r^2} \right)} m = \frac {GM} {r^2}$$

Ahh I see that makes sense, thank you
 
  • #4
Maybe this is what you are asking since you are using lower case g... The gravitational force on a test mass m at a distance r from the center of the Earth is given by GmM/r2, where M is the mass of the earth. At the Earth's surface, the force is given by GmM/R2, where R is the radius of the earth. For small distances from the surface, this equation still holds well and we use g = GM/R2 for the acceleration due to gravity near the Earth's surface, in which case F=mg.

At appreciable distances from the Earth's surface, we have to use r instead of R, hence the other equation.
 

1. How is gravitational field strength calculated?

Gravitational field strength is calculated by dividing the force of gravity on an object by the object's mass. The formula is F = G(m1m2)/r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two interacting objects, and r is the distance between them.

2. What is the unit of measurement for gravitational field strength?

The unit of measurement for gravitational field strength is newtons per kilogram (N/kg). This unit represents the amount of force experienced by a kilogram of mass in a gravitational field.

3. How does distance affect gravitational field strength?

The gravitational field strength decreases as the distance between two objects increases. This is because the force of gravity is inversely proportional to the square of the distance between two objects.

4. What factors affect the gravitational field strength on Earth?

The gravitational field strength on Earth is primarily affected by the mass and radius of the Earth. It also varies slightly depending on altitude and location on the Earth's surface.

5. How is gravitational field strength related to gravitational potential energy?

Gravitational field strength and gravitational potential energy are directly proportional. This means that as the gravitational field strength increases, the gravitational potential energy also increases. This relationship is important in understanding the behavior of objects in gravitational fields.

Similar threads

  • Special and General Relativity
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
740
  • Other Physics Topics
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Other Physics Topics
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
827
  • Introductory Physics Homework Help
Replies
1
Views
202
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Classical Physics
Replies
16
Views
833
Back
Top