Question

At sea level, a body will have minimum weight at
a)Pole
b)Equator
c)${{42}^{0}}$south latitude
d)${{37}^{0}}$north latitude

Answer 1

Hint: The acceleration of gravity at a given altitude depends on the angular velocity and angle of the body’s position with respect to the centre of earth. At the poles, we will have the angle as ninety degrees. At the equator, the angle made by the body with respect to the centre of the earth is zero degrees.

Formulas used:
${{g}^{1}}=g-{{\omega }^{2}}R\cos \alpha $

Complete answer:
Let us assume the gravitational acceleration acting on the body as g. If the radius of earth is R, the angular velocity is $\omega $and it is making an angle $\alpha $with the centre of the earth
Now, for the gravitational acceleration to be minimum,
${{g}^{1}}$must be minimum, i.e., the extra term other than g must be maximum
Therefore,
$\begin{align}
  & \cos \alpha =\max \\
 & \alpha ={{0}^{0}} \\
\end{align}$
The angle between the body and the centre of the earth is zero when the body is at equator.
Therefore, the correct option is option b.

Additional information:
Gravitational acceleration is the free fall acceleration of the object in vacuum that is when there’s no presence of any drag. At a given position on earth, all objects will travel at the same acceleration when present in vacuum. This is true for all masses and compositions of the body. At different points on the earth surface, the free fall acceleration depends on altitude and latitude, with a conventional standard value. In these cases, the other effects such as buoyancy or drag will not be considered.

Note:
The gravitational acceleration acting on the body doesn’t depend on the altitude, but the cosine function will consider the centrifugal force produced by the rotation of the earth. The gravitational acceleration does not depend on the mass of the object or the composition of the object. The gravitational acceleration depends on the angular velocity of the earth and the angle made by the body with the centre of the earth.