Effective Gravity on a Rotating Planet
In the rotating Earth frame the effective gravitational acceleration is the vector sum of true gravity and centrifugal acceleration; at the equator its magnitude is approximately g_eff = g - omega^2 R.
Why this shows up in the exam
Pole-equator weight comparison · Critical angular speed · Latitude effects on apparent gravity
Learn the idea
Rotation reduces apparent weight most at the equator and not at the poles. A spring balance responds to support force, so part of gravity supplies the centripetal acceleration needed to rotate with Earth.
🧠 Memory hook: Rotation borrows from the balance reading.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- g_eff,equator = g - omega² R — equatorial effective gravity for a spherical Earth
- N = m g_eff — spring-balance reading for a stationary surface observer
How to approach it
- 1Identify latitude and rotation radius
- 2Separate true and effective gravity
- 3Use the balance reading, not gravitational force alone
Common slip-ups that cost marks
- •Subtracting omega squared R at the poles
- •Calling apparent weight a change of mass
- •Ignoring that centrifugal acceleration is direction-dependent
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A satellite moves in a circular orbit where GM = 4 x 10^14 m^3/s^2 and orbital radius is 2 x 10^6 m. Find its orbital speed.
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