Gravitation Force on the Earth's Centre

  • Context: Undergrad 
  • Thread starter Thread starter xCross
  • Start date Start date
  • Tags Tags
    Force Gravitation
Click For Summary
SUMMARY

The gravitational force at the Earth's center is zero due to the symmetrical distribution of mass in a uniform sphere. According to Newton's Law of Gravitation, the force is calculated using the formula F=Gm1m2/r^2. However, when considering points inside the sphere, the gravitational force is linearly proportional to the distance from the center, resulting in zero force at the center (r=0). This phenomenon occurs because the gravitational forces exerted by mass on opposite sides of the center cancel each other out.

PREREQUISITES
  • Understanding of Newton's Law of Gravitation
  • Familiarity with spherical mass distributions
  • Basic knowledge of calculus for integration concepts
  • Concept of gravitational force and pressure differentiation
NEXT STEPS
  • Study the derivation of gravitational force inside a uniform sphere
  • Learn about the implications of gravitational symmetry in physics
  • Explore the relationship between gravitational force and pressure at various depths
  • Investigate the concept of gravitational potential energy in spherical bodies
USEFUL FOR

Students of physics, educators teaching gravitational concepts, and anyone interested in understanding the principles of gravitational forces within celestial bodies.

xCross
Messages
15
Reaction score
0
I am having some problem understanding this.. By Newtons Gravitation Law we know:

F=Gm1m2/r^2

centre of the particle means r=0, am I right? if so, the equation says that the force will be infinite on the centre of the earth.. but we see that it has 0 force on the centre.. why such thing happened??
 
Astronomy news on Phys.org
xCross said:
I am having some problem understanding this.. By Newtons Gravitation Law we know:

F=Gm1m2/r^2

centre of the particle means r=0, am I right? if so, the equation says that the force will be infinite on the centre of the earth.. but we see that it has 0 force on the centre.. why such thing happened??
That is the formula for the gravitational force of a point mass which can also be shown, by integrating over the volume of a sphere, to be the force of a spherical mass on any point outside the sphere.

Inside the sphere, if, say r< r0, where r0 is the radius of the sphere, it can be shown that the force for the mass outside distance r cancels. (The mass on "this" side pulls upward, on "that" side pulls downward and so cancel.) That is, only the mass for radius < r gives a non-zero total contribution to the force. Since mass (assuming constant density) is proportional to volume which is proportional to r3, the mass of a sphere of radius r inside a uniform sphere of radius r0 of mass M is Mr3/r03.

Putting that together with GmM/r2, the force on an object at distance r from the center of the sphere is Gm(Mr3/r03)/r2 and so is proportional to r, not 1/r2. Of course, at the center of the Earth r= 0 and so the gravitational force is 0. (Again, because of the symmetry of the sphere and the assumed constant density, the gravitational force from any "bit" of matter in the Earth is canceled by the force of a diametrically opposite "bit" of matter.)

Oh, and that does not mean that the PRESSURE on something at the center of the Earth due to the mass above it would be 0. That's a completely different calculation.
 
Last edited by a moderator:

Similar threads

  • · Replies 84 ·
3
Replies
84
Views
7K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
16
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K