Final answer:
The rotational period of the planet is 6 hours, and the shape of the planet does need to be taken into account.
Step-by-step explanation:
In this question, we are given that the measured weight at the equator is one-half the measured weight at the pole on a planet with the same mass and diameter as Earth. We need to find the rotational period of the planet.
To find the rotational period, we can use the concept of centripetal force. The centripetal force acting on an object depends on its mass, velocity, and the radius of the circular path it follows. Since the planet's mass and diameter are the same as Earth's, the only difference between the equator and the pole is the radius. At the equator, the radius is larger, and at the pole, the radius is smaller. Therefore, the object at the equator will experience a smaller centripetal force compared to the pole.
We know that the weight of an object is the force of gravity acting on it. Therefore, if the weight at the equator is one-half the weight at the pole, the centripetal force at the equator must be one-half the centripetal force at the pole. Since the mass of the object does not change, this means that the velocity at the equator must be higher compared to the pole.
Let's suppose the weight at the pole is W and the weight at the equator is 0.5W. We can use the concept of weight to calculate the centripetal force at the pole and the equator:
At the pole: weight = centripetal force
W = m * g
At the equator: weight = centripetal force
0.5W = m * g'
Here, m is the mass of the object and g and g' are the accelerations due to gravity at the pole and the equator respectively. Since the mass does not change, we can cancel it out from the equations:
At the pole: W / g = centripetal force
At the equator: 0.5W / g' = centripetal force
Since the centripetal force at the equator is one-half the centripetal force at the pole, we can write the ratio of g to g' as:
0.5 = g' / g
g' = 0.5g
Now, we know that the acceleration due to gravity at the pole is equal to the gravitational constant times the mass of the planet divided by the square of the radius at the pole:
g = (GM) / (Rp^2)
Similarly, the acceleration due to gravity at the equator is given by:
g' = (GM) / (Re^2)
where G is the gravitational constant, M is the mass of the planet, Rp is the radius at the pole, and Re is the radius at the equator.
Substituting the values of g and g' in terms of Rp and Re, we have:
(GM) / (Rp^2) = 0.5 * (GM) / (Re^2)
(Rp^2) = 2 * (Re^2)
Rp = sqrt(2) * Re
So, the radius at the pole is equal to the square root of 2 times the radius at the equator.
Now, we know that the circumference of a circle is given by 2π times the radius. Therefore, the distance traveled by an object on the equator during one rotation is 2π times the radius at the equator, and the distance traveled by an object on the pole during one rotation is 2π times the radius at the pole.
Since the object on the equator is traveling a larger distance in the same amount of time compared to the object on the pole, its velocity must be higher.
The time taken for one rotation is the same for both objects and is called the rotational period of the planet. Since the velocity at the equator is higher, the rotational period must be shorter.
Therefore, the correct answer is 6 hours (a).
As for question b, we would need to take the shape of the planet into account. The shape of the planet affects the distribution of mass and therefore the gravitational force acting on objects at different points on the planet's surface. This, in turn, affects the measured weights at the equator and the pole, and ultimately the rotational period of the planet.