Chapter 9, Problem 51SP

Suppose the Earth is a perfect sphere with R = 6370 km. If a person weighs exactly 600.0 N at the North Pole, how much will the person weigh at the equator? [Hint: The upward push of the scale on the person is what the scale will read and is what we are calling the weight in this case.]

To determine

The weight of a person at the equator given that the person weighs $600\text{ N}$ on the North Pole, assuming Earth is a perfect sphere with a radiusof $6370\text{ km}$.

Answer to Problem 51SP

Solution: $5\text{97}\text{.9}\text{N}$

Explanation of Solution

Given data:

The person weighs $600\text{ N}$ on the North Pole.

Consider that Earth is a perfect sphere with a radius of $6370\text{ km}$.

Formula used:

The weight of a body is expressed as,

$W=mg$

Here, $m$ is the mass of the body, $g$ is the acceleration due to gravity, and $W$ is the weight of the body.

The centripetal force on a body is expressed as,

$F_C=mr{\omega }^2$

Here, $r$ is the radius of circle of rotation, $F_C$ is the centripetal force on the body, and $\omega$ is the angular velocity of the body.

The expression for angular velocity is written as,

$\omega =\frac{2\pi }{T}$

Here, $T$ is the time taken for one rotation.

The gravitational force on a body due to Earth is expressed as,

$F_g=\frac{GMm}{R^2}$

Here, $F_g$ is the gravitational force on the body due to Earth, $M$ is the mass of Earth, $R$ is the distance of the body from the centre of Earth, and $G$ is the universal gravitational constant.

The force on the body due to gravity of earth is equal to its weight.

$W=F_g$

Substitute $\frac{GMm}{R^2}$ for $F_g$ and $mg$ for $W$.

$\begin{array}{c}mg=\frac{GMm}{R^2}\\ g=\frac{GM}{R^2}\end{array}$

The expression for acceleration due to gravity in terms of radius of rotation is written as,

$g=\frac{GM}{R^2}$

Explanation:

Consider the expression for acceleration due to gravity in terms of radius of rotation.

$g=\frac{GM}{R^2}$

Understand that the standard values of $G$ and $M$ are $6.67\times {10}^{-11}{\text{ m}}^3/\left(\text{kg}\cdot {\text{s}}^2\right)$ and $6\times {10}^{24}\text{ kg}$, respectively. Therefore, substitute $6.67\times {10}^{-11}{\text{ m}}^3/\left(\text{kg}\cdot {\text{s}}^2\right)$ for $G$, $6370\text{ km}$ for $R$, and the standard value of mass of Earth, that is, $6\times {10}^{24}\text{ kg}$ for $M$.

$\begin{array}{c}g=\frac{\left(6.67\times {10}^{-11}{\text{ m}}^3/\left(\text{kg}\cdot {\text{s}}^2\right)\right)\left(6\times {10}^{24}\text{ kg}\right)}{{\left(6370\text{ km}\right)}^2}\\ =\frac{\left(6.67\times {10}^{-11}{\text{ m}}^3/\left(\text{kg}\cdot {\text{s}}^2\right)\right)\left(6\times {10}^{24}\text{ kg}\right)}{\left(40576900{\text{ km}}^2\right)}\\ =\frac{\left(6.67\times {10}^{-11}{\text{ m}}^3/\left(\text{kg}\cdot {\text{s}}^2\right)\right)\left(6\times {10}^{24}\text{ kg}\right)}{\left(40576900{\text{ km}}^2\right)\left(\frac{{10}^6{\text{ m}}^2}{1{\text{km}}^2}\right)}\\ =9.86{\text{ m/s}}^2\end{array}$

Understand that the North Pole lies on the axis of rotation of the Earth. Therefore, the distance from the axis of Earth is zero. Therefore, the weight of the body on North Pole is only due to the gravitational force.

Consider the expression for weight of the body at North Pole.

$W_1=m_{1}g$

Here, $m_1$ is the mass of the personand $W_1$ is the weight of the person at North Pole.

Substitute $9.86{\text{ m/s}}^2$ for $g$ and $600\text{ N}$ for $W_1$

$\begin{array}{c}600\text{ N}=m_1\left(9.86{\text{ m/s}}^2\right)\\ m_1=\frac{600\text{ N}}{9.86{\text{ m/s}}^2}\\ m_1=60.85\text{ kg}\end{array}$

Calculate the angular speed of Earth.

$\omega =\frac{2\pi }{T}$

Understand that time period of rotation of Earth is $24\text{hours}$. Therefore, substitute $24\text{hours}$ for $T$.

$\begin{array}{c}\omega =\frac{2\pi }{24\text{hours}}\\ =\frac{2\pi }{24\text{hours}}\left(\frac{1\text{hour}}{3600\text{ s}}\right)\\ =7.272\times {10}^{-5}\text{ rad/s}\end{array}$

Understand that when the person is at the equator, the person is at a distance equal to radius of Earth from the axis of rotation of the Earth. Therefore, the weight of the body on equator is due to the difference of gravitational and centripetal force.

Consider the expression for weight of the person at equator.

$W_2=W_1-F_{C1}$

Here, $W_2$ is the weight of the person at equator and $F_{C1}$ is the centripetal force on the person at equator.

Substitute $m_{1}g$ for $W_1$ and $m_{1}r{\omega }^2$ for $F_C$

$W_2=m_{1}g-m_{1}r{\omega }^2$

Substitute $7.272\times {10}^{-5}\text{ rad/s}$ for $\omega$, $9.86{\text{ m/s}}^2$ for $g$, $60.85\text{ kg}$ for $m_1$, and $6370\text{ km}$ for $r$

$\begin{array}{c}{W}_2=\left(60.85\text{ kg}\right)\left(9.86{\text{ m/s}}^2\right)-\left(60.85\text{ kg}\right)\left(6370\text{ km}\right){\left(7.272\times {10}^{-5}\text{ rad/s}\right)}^2\\ =\left(60.85\text{ kg}\right)\left(9.86{\text{ m/s}}^2\right)-\left(60.85\text{ kg}\right)\left(6370\text{ km}\right)\left(\frac{1000\text{ m}}{1\text{km}}\right)\left(5.288\times {10}^{-9}{\text{ rad}}^2{\text{/s}}^2\right)\\ =599.98\text{N}-2.05\text{N}\\ =5\text{97}\text{.9}\text{N}\end{array}$

Conclusion:

The weight of the person at equator is $5\text{97}\text{.9}\text{N}$.