Run Fastest to Weigh Most at Equator!

  • Thread starter Thread starter Johny158
  • Start date Start date
  • Tags Tags
    Equator
AI Thread Summary
To weigh the most while running at the equator, one must consider the effects of centrifugal force due to the Earth's rotation. Gravitational acceleration is slightly lower at the equator (9.78 m/s²) compared to the poles (9.83 m/s²), resulting in a weight difference. The net force acting on a person running at the equator is influenced by this centrifugal effect, which can be minimized by running in the direction opposite to the Earth's rotation. While the gravitational force is nearly equal at both locations, the apparent weight is affected by the Earth's shape and rotation. Understanding these dynamics is crucial for determining how speed impacts weight at the equator.
Johny158
Messages
1
Reaction score
0

Homework Statement


What speed you have to run on the equator to weigh as much as possible (the greatest weight) if you do
choose the right direction? Do not consider relativity.

Homework Equations


ravitational acceleration is 9.78 m/s2 at the equator and 9.83 m/s2 at the poles, so you weigh about 0.5% more at the poles than at the equator.

The Attempt at a Solution


Absolutely don't know :D
 
Physics news on Phys.org
Johny158 said:
ravitational acceleration is 9.78 m/s2 at the equator and 9.83 m/s2 at the poles, so you weigh about 0.5% more at the poles than at the equator.

That's not really relevant to this question. This isn't about the difference between the poles and equator.

Sorry I should correct my initial reply. Gravitational force is actually about the same at the equator and the poles. The net force acting on you is different at the equator and poles.

Hint: It's about an effect that occurs at all latitudes except the poles and how you could reduce this effect?

Hintt 2: The Earth isn't a perfect sphere. Why?
 
CWatters said:
Gravitational force is actually about the same at the equator and the poles.
Well, it is about 0.2% less at the equator, but as you say the apparent gravitational force (the ratio between weight and mass) is about 0.5% less.
 
  • Like
Likes CWatters
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top