Question

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of $\mathbf{0}\mathbf{.}\mathbf{0337}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$, whereas a point at the poles experiences no centripetal acceleration. If a person at the equator has a mass of$\mathbf{75}\mathbf{.}\mathbf{0}\mathit{k}\mathit{g}$, calculate (a) the gravitational force (true weight) on the person and (b) the normal force (apparent weight) on the person. (c) Which force is greater? Assume the Earth is a uniform sphere and take$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{800}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$.

Short answer

(a) The true weight of the person is $735N$.

(b) The apparent weight of the person is $732N$.

(c) The gravitational force is greater than the normal force.

Step-by-step solution

Defining gravitational force

Gravitational force is the force exerted by the earth on object. This force is directed towards the centre of the earth.

The expression for the gravitational force is given by

$F=mg$

Here$F$ is the gravitational force,$m$ is the mass of the body and $g$is the acceleration due to gravity.

Calculating the true weight

(a)

The gravitational force exerted by the earth on the person is

$W=mg$

Substitute 75 kg for m and $9.8m/s^2$for g in the above equation.

$$\begin{gathered} W=\left(75\right)\left(9.8\right) \\ =735 \text{N} \end{gathered}$$

Therefore, the true weight of the person is $735N$.

Calculating the apparent weight

(b)

The apparent weight of the person using Newton’s second law is given by

$W_{app}=mg-ma$

Here m is the mass of the person, g is the acceleration due to gravity and $a$is the centripetal acceleration at equator.

Substitute $75kg$for $m$and $9.8m/s^2$for g and $0.037m/s^2$for $a$in the above equation.

$$\begin{gathered} W_{app}=75\left(9.8\right)-\left(75\right)\left(0.0337\right) \\ =732 \text{N} \end{gathered}$$

Therefore, the apparent weight of the person is $732N$.

Explanation for part (c)

(c)

From the above calculation of part (a) and (b), it is clear that the gravitational force is greater than the nornal force.