Calculating g with centripetal acceleration

Click For Summary
SUMMARY

The discussion centers on calculating gravitational acceleration (g) using the formula a = w² * r, where w is angular velocity and r is the radius of the Earth. The user initially calculated the angular velocity as 2π/86400 radians per second and used the Earth's radius of 6,378,100 meters, resulting in a centripetal acceleration of 0.0337 m/s². However, this value does not represent g, as it reflects the centripetal acceleration due to Earth's rotation rather than gravitational acceleration. The correct formula for calculating g at the Earth's surface is g = GM/R², where G is the gravitational constant and M is the Earth's mass.

PREREQUISITES
  • Understanding of angular velocity and its conversion to radians per second
  • Familiarity with centripetal acceleration and its calculation
  • Knowledge of gravitational force and the formula g = GM/R²
  • Basic concepts of forces acting on objects at rest on Earth's surface
NEXT STEPS
  • Study the derivation and applications of the formula g = GM/R²
  • Learn about the differences between centripetal and centrifugal forces
  • Explore the effects of Earth's rotation on gravitational measurements
  • Investigate the implications of gravitational acceleration in different contexts, such as at the equator versus the poles
USEFUL FOR

Physics students, educators, and anyone interested in understanding the principles of gravitational acceleration and the effects of Earth's rotation on objects at its surface.

dmayers94
Messages
16
Reaction score
0
Hi, I'm trying to find g, or 9.8 m/s^2, with the formula a = w^2 * r. First, I found the value for the angular velocity (2*pi/24 hours) and i converted this to radians per second finding the value 2*pi/86400. I googled the radius for the Earth and got 6,378,100 meters. Finally, i plugged these values into the formula to obtain a centripetal acceleration of 0.0337 m/s^2 which obviously isn't the answer I was looking for. Can someone find the holes in my math or logic? Thanks.
 
Physics news on Phys.org
I haven't checked the numbers, but that formula won't give you g anyway. It will instead give you the net acceleration required to keep an object on the Earth's surface moving in the circular path given by the Earth's radius. At the Earth's surface, both the gravitational force and the normal force of the ground act on objects. What you have found is the difference between the accelerations caused by these two forces.

If gravity were the only force acting on an object at the Earth's surface, it would be pulled down into a much tighter "orbit." One with a radius much smaller than earth's. But this is not possible since contact forces with the ground counteract gravity somewhat, and they balance so that we stay on the surface of the Earth at all times.
 
This formula is not supposed to give you g.

you get g at a distance R from the Earth's centre using g=GM/R^2

where M is mass of earth
G is the Universal Gravitation constant.
For g at surface we use R as radius of earth.

What you are calculating by a=w^2*R is something entirely else.

You have calculated the centripetal acceleration felt by a particle moving in a circle of radius R at the same angular frequency as that of earth.

That has no reason to be g.

Infact its the centripetal acceleration a person on equator will have wrt to the Centre of Mass of Earth (but we don't feel as we are at rest relative to Earth's surface and because its too small) cause we are rotating with earth.
 
Last edited:
Gravity is caused by masses attracting each other and not because they are spinning. Non-spinning objects attract each other just as well as spinning objects.
 

Similar threads

  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 43 ·
2
Replies
43
Views
11K
Replies
12
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K