- #1

lpettigrew

- 115

- 10

- Homework Statement
- Hello, I have found the following question which appeared to be quite straightforward but the more I think about it the more I complicate matters. If the measured value of gravity, g, at the equator is 9.78 how much g be if the Earth was not rotating?

I have tried to answer this fully and give reasoning to my solution but I have struggled to calculate the correct value of the centripetal force acting. I would be very grateful for any further explanation or advice

- Relevant Equations
- F=mv^2/r

v=2πr/T

If we think about the forces acting on an object at the surface of the Earth, these forces would be the weight force (mg) pulling downwards and the normal force (N) acting upwards . If the Earth were to stop rotating, then according to Newton's first law the forces must balance, meaning the net force is zero.

However, when the Earth is rotating an object on the surface of the Earth would experience an acceleration, supplied by the net force acting, the centripetal force pointing toward the centre of the Earth.

The centripetal force can be calculated as F=mv^2/r.

Hence, ΣFnet=Fc=Fg-FN=mg-N

To find the value of g if the Earth were still would this be the apparent weight of the object, i.e the normal force?

N=mg-Fc=mg-mv^2/r

g'=mg-mv^2/r

I think mass could cancel here;

g'=g-v^2/r

If g is measured to be 9.78 at the equator when the Earth is spinning then;

g'=9.78-465^2/6.4*10^6

g'=9.74621 ~9.75 ms^-2

This would mean ∆g =9.78-9.75~0.03ms^-2

Or alternatively finding the centripetal force one could subtract this directly from the measured value of g at 9.78ms^-2;

Take v=2πr/T

T=86400s

r=6.4*10^6m

Thus, v=2*π*6.4*10^6/86400

v=465.4211..~465 ms^-1

Then, the force acting on an object of 1kg: F=1*465^2/6.4*10^6=0.0337..~0.034 N

However, this is my point of difficulty, since I have seen elsewhere that the centripetal force acting on an object at the equator is equal to 0.34N, how am I off by a factor of 10? Have I found the centripetal acceleration instead?

However, when the Earth is rotating an object on the surface of the Earth would experience an acceleration, supplied by the net force acting, the centripetal force pointing toward the centre of the Earth.

The centripetal force can be calculated as F=mv^2/r.

Hence, ΣFnet=Fc=Fg-FN=mg-N

To find the value of g if the Earth were still would this be the apparent weight of the object, i.e the normal force?

N=mg-Fc=mg-mv^2/r

g'=mg-mv^2/r

I think mass could cancel here;

g'=g-v^2/r

If g is measured to be 9.78 at the equator when the Earth is spinning then;

g'=9.78-465^2/6.4*10^6

g'=9.74621 ~9.75 ms^-2

This would mean ∆g =9.78-9.75~0.03ms^-2

Or alternatively finding the centripetal force one could subtract this directly from the measured value of g at 9.78ms^-2;

Take v=2πr/T

T=86400s

r=6.4*10^6m

Thus, v=2*π*6.4*10^6/86400

v=465.4211..~465 ms^-1

Then, the force acting on an object of 1kg: F=1*465^2/6.4*10^6=0.0337..~0.034 N

However, this is my point of difficulty, since I have seen elsewhere that the centripetal force acting on an object at the equator is equal to 0.34N, how am I off by a factor of 10? Have I found the centripetal acceleration instead?