Question

A planet similar to the Earth has a radius 7.4 × 106 m and has an acceleration of gravity of 10 m/s 2 on the planet’s surface. The planet rotates about its axis with a period of 25 h. Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, what is the new period? Answer in units of h.
Part 2:
By what factor would the speed of the object be increased when the planet is rotating at the higher speed?

Answer 1

Answer:

1.5 hr

16.7

Explanation:

Zero apparent weight means there's no normal force.

Sum the forces in the centripetal direction.

∑F = ma

mg = mv²/r

v = √(gr)

v = √(7.4×10⁶ m × 10 m/s²)

v = 8602 m/s

The circumference of the equator is:

C = 2πr

C = 2π (7.4×10⁶ m)

C = 4.65×10⁷ m

So the period is:

T = C / v

T = (4.65×10⁷ m) / (8602 m/s)

T = 5405 s

T = 1.5 hr

The initial speed is:

v = C / T

v = (4.65×10⁷ m) / (25 h × 3600 s/h)

v = 517 m/s

The speed increases by a factor of:

8602 m/s / 517 m/s

16.7