Chapter 5, Problem 39P

a)

To determine

The value of acceleration due to gravity at 6400 m above theEarth ’s surface.

Answer to Problem 39P

Solution:

The value of acceleration due to gravity at above theEarth ’s surface is $9.74{\text{m/s}}^2$.

Explanation of Solution

Given:

Height =6400 m

Formula Used:

Acceleration due to gravity, g :

  $g=G\frac{M}{r^2}$

M is the mass of the planet, G is the gravitational constant, andris thedistanceof the point from the planet where g is required to be found.

Calculation:

  $\begin{array}{c}r=6.38\times {10}^6\text{m}+6400\text{m}\\ \text{=6}\text{.386}\times {10}^6\text{m}\end{array}$

Substitute the values and solve:

  $\begin{array}{l}{g}_{\text{Earth}}=G\frac{M_{\text{Earth}}}{r_{\text{Earth}}^{{}^2}}\\ =\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(6.386\times {10}^6\text{m}\right)}^2}\\ =9.74{\text{m/s}}^2\end{array}$

Conclusion:

Therefore, the value of acceleration due to gravity at above theEarth’s surface is $9.74{\text{m/s}}^2$

b)

To determine

The value of acceleration due to gravity at 6400 kmabove theEarth ’s surface.

Answer to Problem 39P

Solution:

The value of acceleration due to gravity at above theEarth ’s surface is $2.44{\text{m/s}}^2$.

Explanation of Solution

Given:

Height =6400 km

Formula Used:

Acceleration due to gravity, g :

  $g=G\frac{M}{r^2}$

M is the mass of the planet, G is the gravitational constant, and r is the distanceof the point from the planet where g is required to be found.

Calculation:

  $\begin{array}{c}r=6.38\times {10}^6\text{m}+6400\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\\ \text{=6}\text{.386}\times {10}^6\text{m+6}\text{.4}\times {\text{10}}^6\text{m}\\ \text{=1}\text{.278}\times {\text{10}}^7\text{m}\end{array}$

Substitute the values and solve:

  $\begin{array}{l}{g}_{\text{Earth}}=G\frac{M_{\text{Earth}}}{r_{\text{Earth}}^{{}^2}}\\ =\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(1.278\times {10}^7\text{m}\right)}^2}\\ =2.44{\text{m/s}}^2\end{array}$

Conclusion:

Therefore, the value of acceleration due to gravity at above theEarth’s surface is $2.44{\text{m/s}}^2$.