What Is the New Rotation Period for Zero Apparent Weight at the Equator?

[muahe]

Homework Statement

A planet similar to the Earth has a radius 4.4*10^6m 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 25h. Imagine that the rotational speed can be increased.
a/If an object at the equator is to have zero apparent weight, what is the new period?
b/By what factor would the speed of the object be increased when the planet is rotating at the higher speed?

Homework Equations

The Attempt at a Solution

I found the Velocity of the planet = 6633.25, but then i don't know what do to next. I don't understand what does " obj. has zero apparent weight" mean?
b/ i have no idea about it...

 

[quantumdude]

Regarding "zero apparent weight": Imagine you are standing on a scale. What the scale is actually reading is not your weight, but rather the normal force exerted on you by the scale. Draw a free body diagram and set up Newton's second law, and then ask yourself under what condition would that normal force drop to zero.

 

[Moetasim]

I can provide some insights and explanations on circular motion of a planet and address the questions raised in the content.

Circular motion of a planet is a result of the gravitational force exerted by its star, which keeps the planet in a stable orbit. In this case, the planet's radius and acceleration of gravity are given as 4.4*10^6m and 10 m/s^2 respectively. This means that at the planet's surface, an object with a mass of 1kg would experience an apparent weight of 10 N.

Now, if we increase the rotational speed of the planet, the centrifugal force acting on the object at the equator would also increase. This force would oppose the gravitational force, resulting in a decrease in the apparent weight of the object. When the apparent weight becomes zero, it means that the centrifugal force is equal to the gravitational force, and the object is essentially in a state of freefall.

To calculate the new period of rotation, we can use the equation T=2πr/v, where T is the period of rotation, r is the radius of the planet, and v is the velocity of rotation. Since the velocity of the planet is directly proportional to its rotational speed, the new period would also decrease.

For part b, we can use the equation v=ωr, where ω is the angular speed of rotation. As the rotational speed increases, the angular speed would also increase, resulting in a higher velocity. The factor by which the speed would increase can be calculated by dividing the new velocity by the initial velocity.

In conclusion, circular motion of a planet is a complex phenomenon that is affected by multiple factors, including the planet's radius, acceleration of gravity, and rotational speed. By understanding these concepts and using relevant equations, we can make predictions and calculations about the behavior of a planet in different scenarios.

 

[ogb p]

I would like to clarify that the statement "an object at the equator has zero apparent weight" is not entirely accurate. All objects have weight, which is the force of gravity acting on them. What the statement is likely referring to is the concept of weightlessness, which is a state in which an object does not experience the sensation of weight due to being in a state of freefall or orbit.

To calculate the new period of the planet's rotation, we can use the formula T = 2π√(R/g), where T is the period, R is the radius, and g is the acceleration of gravity. Plugging in the given values, we get T = 2π√(4.4*10^6/10) = 2π*2099.02 seconds. If the rotational speed is increased, the new period will be shorter, but the exact value cannot be determined without knowing the new velocity.

To determine the factor by which the speed of the object would be increased, we can use the formula v = ωR, where v is the velocity, ω is the angular velocity, and R is the radius. Since the radius remains constant, the speed will increase in direct proportion to the angular velocity. Therefore, if the angular velocity is doubled, the speed will be doubled as well.