Chapter 10, Problem 28QAP

To determine

(a)

The gravitational force on the passenger at cruising altitude.

Answer to Problem 28QAP

Gravitational force on the passenger at cruising altitude is 683.97 N

Explanation of Solution

Given:

Mass of earth, $m_{\text{Earth}}=5.98\times {10}^{24}kg$

The average mass of the person, $m_{Person}=70kg$

Cruising height of flight, h = 10000 m

Radius of earth, $R_{Earth}=6.38\times {10}^{6}m$

Formula used:

The gravitational force on the passenger

  $F=\frac{G{m}_{Person}{m}_{\text{ Earth }}}{{\left(R_{earth}+h\right)}_{}^2}$

Where,

h = height of flight from sea level.

Calculation:

The gravitational force on the passenger at cruising height ( h = 10⁴m)

  $F=\frac{G{m}_{Person}{m}_{\text{ Earth }}}{{\left(R_{earth}+h\right)}_{}^2}=\frac{\left(6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^2}{{\text{kg}}^2}\right)\left(\text{70}\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(6.38\times {10}^6\text{m}\text{+}1{\text{0}}^4\text{m}\right)}^2}=683.79N$

Conclusion:

Gravitational force on the passenger at cruising altitude is 683.97 N

To determine

(b)

How much "lighter" would a person be in flight as compared to at sea level?

Answer to Problem 28QAP

Person become 2.14 N lighter in flight as compared to at sea level.

Explanation of Solution

Given:

Mass of earth, $m_{\text{Earth}}=5.98\times {10}^{24}kg$

The average mass of the person, $m_{Person}=70kg$

Cruising height of flight, h = 10000 m

Radius of earth, $R_{Earth}=6.38\times {10}^{6}m$

Formula used:

The gravitational force on the passenger

  $F=\frac{G{m}_{Person}{m}_{\text{ Earth }}}{{\left(R_{earth}+h\right)}_{}^2}$

Where,

h = height of flight from sea level.

Calculation:

Weight of the person above the earth surface is equal to the magnitude of gravitational force acting on that person.

  $W=\left|F\right|=\left|\frac{G{m}_{Person}{m}_{\text{ Earth }}}{{\left(R_{earth}+h\right)}_{}^2}\right|$

The weight of person at sea level (h = 0)

  $W_{\text{sea}\text{level}}=\left|\frac{G{m}_{Person}{m}_{\text{ Earth }}}{{\left(R_{earth}+h\right)}_{}^2}\right|=\left|\frac{\left(6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^2}{{\text{kg}}^2}\right)\left(\text{70}\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(6.38\times {10}^6\text{m}\text{+}\text{0}\text{m}\right)}^2}\right|=685.93N$

The weight of person at cruising height ( h = 10⁴m)

  $W_{\text{cruising}\text{height}}=\left|\frac{G{m}_{Person}{m}_{\text{ Earth }}}{{\left(R_{earth}+h\right)}_{}^2}\right|=\left|\frac{\left(6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^2}{{\text{kg}}^2}\right)\left(\text{70}\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(6.38\times {10}^6\text{m}\text{+}1{\text{0}}^4\text{m}\right)}^2}\right|=683.79N$

Change in weight,

  $W_{\text{sea}\text{level}}-W_{\text{cruising}\text{height}}=\left(685.93N\right)-\left(683.79N\right)=2.14N$

Conclusion:

Person become 2.14 N lighter in flight as compared to at sea level.