Chapter 7, Problem 65A

a.

To determine

Weight of spacecraft at given height.

Answer to Problem 65A

Weight of spacecraft is $1.8\times {10}^3\text{N}$ .

Explanation of Solution

Given:

Radius of earth $r_E=6.38\times {10}^3\text{km}$

Weight of spacecraft is $F_g=7.2\times {10}^3\text{N}$

Height of spacecraft is $h=6.38\times {10}^3\text{km}$

Formula used:

The acceleration due to gravity at height h from earth’s surface is,

  $g_h=g{\left(1+\frac{h}{r_E}\right)}^2$

Where, $h$ is height from earth’s surface, $r_E$ is radius of earth and $g$ is acceleration due to gravity.

Calculation:

On substituting the given values,

  $\begin{array}{l}{g}_h=\frac{g}{{\left(1+\frac{h}{r_E}\right)}^2}\\ g_h=\frac{g}{{\left(1+\frac{6.38\times {10}^3}{6.38\times {10}^3}\right)}^2}\\ \Rightarrow g_h=\frac{g}{4}\end{array}$

So,

  $\begin{array}{c}{F}_g\text{'}=m{g}_h\\ =m\frac{g}{4}\\ =\frac{F_g}{4}\\ =\frac{1}{4}\times 7.2\times {10}^3\text{N}\\ =1.8\times {10}^3\text{N}\end{array}$

Conclusion:

Therefore, the weight of the spacecraft is $1.8\times {10}^3\text{N}$ .

b.

To determine

Weight of spacecraft at given distance.

Answer to Problem 65A

Weight of spacecraft is $8\times {10}^2\text{N}$ .

Explanation of Solution

Given:

Radius of earth $r_E=6.38\times {10}^3\text{km}$

Weight of spacecraft is $F_g=7.2\times {10}^3\text{N}$

Height of spacecraft is $d=1.28\times {10}^4\text{km}$

Formula used:

The acceleration due to gravity at height h from earth’s surface is,

  $g_h=g{\left(1+\frac{h}{r_E}\right)}^2$

Where, $h$ is height from earth’s surface, $r_E$ is radius of earth and $g$ is acceleration due to gravity.

Calculation:

On substituting the given values,

  $\begin{array}{l}{g}_h=\frac{g}{{\left(1+\frac{h}{r_E}\right)}^2}\\ g_h=\frac{g}{{\left(1+\frac{1.28\times {10}^4}{6.38\times {10}^3}\right)}^2}\\ \Rightarrow g_h=\frac{g}{9}\end{array}$

So,

  $\begin{array}{c}{F}_g\text{'}=m{g}_h\\ =m\frac{g}{9}\\ =\frac{F_g}{9}\\ =\frac{1}{9}\times 7.2\times {10}^3\text{N}\\ =8\times {10}^2\text{N}\end{array}$

Conclusion:

Therefore, the weight of the spacecraft is $1.8\times {10}^3\text{N}$ .

c.

To determine

Weight of spacecraft at given distance.

Answer to Problem 65A

Weight of spacecraft is $2.92\times {10}^2\text{N}$ .

Explanation of Solution

Given:

Radius of earth $r_E=6.38\times {10}^3\text{km}$

Weight of spacecraft is $F_g=7.2\times {10}^3\text{N}$

Height of spacecraft is $d=2.55\times {10}^4\text{km}$

Formula used:

The acceleration due to gravity at height h from earth’s surface is,

  $g_h=g{\left(1+\frac{h}{r_E}\right)}^2$

Where, $h$ is height from earth’s surface, $r_E$ is radius of earth and $g$ is acceleration due to gravity.

Calculation:

On substituting the given values,

  $\begin{array}{l}{g}_h=\frac{g}{{\left(1+\frac{h}{r_E}\right)}^2}\\ g_h=\frac{g}{{\left(1+\frac{2.55\times {10}^4}{6.38\times {10}^3}\right)}^2}\\ \Rightarrow g_h=\frac{g}{25}\end{array}$

So,

  $\begin{array}{c}{F}_g\text{'}=m{g}_h\\ =m\frac{g}{25}\\ =\frac{F_g}{25}\\ =\frac{1}{25}\times 7.2\times {10}^3\text{N}\\ =2.92\times {10}^2\text{N}\end{array}$

Conclusion:

Therefore, the weight of the spacecraft is $2.92\times {10}^2\text{N}$ .