Chapter 5, Problem 85P

(a)

To determine

The magnitude of height H.

Answer to Problem 85P

The magnitude of height H is $3.58\times {10}^7\text{m}$.

Explanation of Solution

The height of the space station is H and radius of earth is $6.371\times {10}^6\text{m}$, mass of the earth is $5.974\times {10}^{24}\text{kg}$, the time period the space station is $86400\text{s}$ and universal gravitational constant is $6.674\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}$.

Write the expression for the radius of the space station.

$R=R_E+H$ (I)

Here, $R_E$ is the radius of the earth, R is the radius of the space station and H is the height of the space station above the surface of the earth.

Write the expression to calculate the radius of the space station.

$R={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{\frac{1}{3}}$ (II)

Here, G is the universal gravitational constant, M is the mass of the earth, T is the time period of space station.

Equate equation (I) and (II) to calculate H.

$R_E+H={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{\frac{1}{3}}$

Substitute $6.371\times {10}^6\text{m}$ for $R_E$, $5.974\times {10}^{24}\text{kg}$ for M, $86400\text{s}$ for T and $6.674\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}$ for G in the above equation to calculate H.

$\begin{array}{c}6.371\times {10}^6\text{m}+H={\left(\frac{\left(6.674\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(5.974\times {10}^{24}\text{kg}\right){\left(86400\text{s}\right)}^2}{4{\pi }^2}\right)}^{\frac{1}{3}}\\ H=4.22\times {10}^7\text{m}-6.371\times {10}^6\text{m}\\ =3.58\times {10}^7\text{m}\end{array}$

Conclusion:

Therefore, the magnitude of height H is $3.58\times {10}^7\text{m}$.

(b)

To determine

The tension on the spring and which part of the cable would experience this tension.

Answer to Problem 85P

The tension in the cable is $\text{55}\text{N}$ and the upward potion of the cable above the car would under tension.

Explanation of Solution

The mass of the car is $100\text{kg}$.

Write the equilibrium condition for forces in the cable.

$\frac{GMm}{r^2}-T_c=m{\omega }^{2}r$ (III)

Here, $T_c$ is the tension in the cable, r is the distance from the centre of the earth to the car, m is the mass of the car and $\omega$ is the angular speed.

Write the expression to calculate the distance from the centre of the earth to the car.

$r=R_E+\frac{H}{2}$ (IV)

Write the expression to calculate the angular speed of the car.

$\omega =\frac{2\pi }{T}$ (V)

Rewrite the equation (III) using (IV) and (V) in terms of T.

$\begin{array}{c}\frac{GMm}{{\left(R_E+\frac{H}{2}\right)}^2}-T_c=m{\left(\frac{2\pi }{T}\right)}^2\left(R_E+\frac{H}{2}\right)\\ T_c=\frac{GMm}{{\left(R_E+\frac{H}{2}\right)}^2}-m{\left(\frac{2\pi }{T}\right)}^2\left(R_E+\frac{H}{2}\right)\end{array}$

Substitute $3.58\times {10}^7\text{m}$ for H, $6.371\times {10}^6\text{m}$ for $R_E$, $5.974\times {10}^{24}\text{kg}$ for M, $100\text{kg}$ for m and $86400\text{s}$ for T, and $6.674\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}$ for G in the above equation to calculate $T_c$.

$\begin{array}{c}{T}_c=\frac{\left(6.674\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(5.974\times {10}^{24}\text{kg}\right)100\text{kg}}{{\left(6.371\times {10}^6\text{m}+\frac{3.58\times {10}^7\text{m}}{2}\right)}^2}-100\text{kg}{\left(\frac{2\pi }{86400\text{s}}\right)}^2\left(6.371\times {10}^6\text{m}+\frac{3.58\times {10}^7\text{m}}{2}\right)\\ =\frac{3.987\times {10}^{16}\text{kg}\cdot {\text{m}}^{\text{3}}/{\text{s}}^{\text{2}}}{5.89\times {10}^{14}{\text{m}}^2}-\left(5.28\times {10}^{-7}\text{kg}/{\text{s}}^{\text{2}}\right)\left(2.43\times {10}^7\text{m}\right)\\ =67.7\text{N}-12.8\text{N}\\ =\text{54}\text{.9}\text{N}\\ \sim \text{55}\text{N}\end{array}$

Since the tension on the car is positive in magnitude, which means car would pull upward. Therefore, the cable segment above the car upward portion would under tension.

Conclusion:

Therefore, the tension in the cable is $\text{55}\text{N}$ and the upward potion of the cable above the car would under tension.