Chapter 6, Problem 55P

Part (a)

To determine

Maximum height of lunar module to shut off the engine if the velocity of the lander relative to the surface at that moment is 0 m/s.

Answer to Problem 55P

Solution:2.8 m

Explanation of Solution

Given:Acceleration due to gravity at moon $\text{=1}\text{.62}{\text{m/s}}^{\text{2}}$

Velocity of the lander relative to the surface at that moment is 0 m/s.

Formula Used:

During landing, only gravitational force due moon is acting conservation of energy.

  $K.E_1+P.E_1=K.E_2+P.E_2$

Calculation:

Initial kinetic energy:

  $\begin{array}{c}\left(K.E_1\right)=\frac{1}{2}m{v}_1{}^2\\ \text{ =}\frac{\text{1}}{\text{2}}\text{×}m{\left(\text{0}\right)}^{\text{2}}\\ =0\end{array}$

Initial potential energy:

  $\begin{array}{c}\left(P.E_1\right)=mg{h}_1\\ =m\text{×1}\text{.62×}{h}_1\end{array}$

Final kinetic energy:

  $\begin{array}{c}\left(K.E_2\right)=\frac{1}{2}m{v}_2{}^2\\ =\frac{1}{2}\times m\times v_2{}^2\\ =\frac{\text{1}}{\text{2}}\text{×}m\text{×}{\left(\text{3}\right)}^{\text{2}}\end{array}$

Final potential energy:

  $\begin{array}{c}\left(P.E_2\right)=mg{h}_2\\ =m\text{×1}\text{.62×}\left(\text{0}\right)\\ \text{=}\text{0}\end{array}$

Apply conservation of energy

  $K.E_1+P.E_1=K.E_2+P.E_2$

  $\begin{array}{l}\text{0+}m\times \text{1}\text{.62}\times h_1\text{=}\frac{\text{1}}{\text{2}}m\times {\left(\text{3}\right)}^{\text{2}}\text{+0}\\ h_1\text{=}\frac{{\left(\text{3}\right)}^{\text{2}}}{\text{2×1}\text{.62}}\\ h_1\approx \text{2}\text{.8}\end{array}$

Conclusion:Maximum height of lunar module to shut off the engine if the velocity of the lander relative to the surface at that moment is 0 m/s is 2.8 m.

Part (b)

To determine

Maximum height of lunar module to shut off the engine if the velocity of the lander relative to the surface at that moment is $2.0$ m/s downward.

Answer to Problem 55P

Solution: 1.5 m

Explanation of Solution

Given: Acceleration due to gravity at moon $\text{=1}\text{.62}{\text{m/s}}^{\text{2}}$

Velocity of the lander relative to the surface at that moment is 2 m/s downwards

Formula Used:

During landing, only gravitational force due moon is acting conservation of energy.

  $K.E_1+P.E_1=K.E_2+P.E_2$

Calculation:

Initial kinetic energy:

  $\begin{array}{c}\left(K.E_1\right)=\frac{1}{2}m{v}_1{}^2\\ \text{ =}\frac{\text{1}}{\text{2}}\text{×}m{\left(2\right)}^{\text{2}}\end{array}$

Initial potential energy:

  $\begin{array}{c}\left(P.E_1\right)=mg{h}_1\\ =m\text{×1}\text{.62×}{h}_1\end{array}$

Final kinetic energy:

  $\begin{array}{c}\left(K.E_2\right)=\frac{1}{2}m{v}_2{}^2\\ =\frac{1}{2}\times m\times v_2{}^2\\ =\frac{\text{1}}{\text{2}}\text{×}m\text{×}{\left(\text{3}\right)}^{\text{2}}\end{array}$

Final potential energy:

  $\begin{array}{c}\left(P.E_2\right)=mg{h}_2\\ =m\text{×1}\text{.62×}\left(\text{0}\right)\\ \text{=}\text{0}\end{array}$

Apply conservation of energy

  $K.E_1+P.E_1=K.E_2+P.E_2$

  $\begin{array}{l}\frac{\text{1}}{\text{2}}m{\left(\text{2}\right)}^{\text{2}}\text{+}m\text{×1}\text{.62×}{h}_1\text{=}\frac{\text{1}}{\text{2}}\text{×}m\text{×}{\left(\text{3}\text{.0}\right)}^{\text{2}}\text{+0}\\ \text{1}\text{.62×}{h}_1\text{=}\frac{\text{1}}{\text{2}}\left({\left(\text{3}\text{.0}\right)}^{\text{2}}\text{-}{\left(\text{2}\text{.0}\right)}^{\text{2}}\right)\\ h_1\text{=}\frac{\frac{\text{1}}{\text{2}}\left[{\left(\text{3}\text{.0}\right)}^{\text{2}}\text{-}{\left(\text{2}\text{.0}\right)}^{\text{2}}\right]}{\text{1}\text{.62}}\\ \text{ =1}\text{.5}\text{m}\end{array}$

Conclusion:Maximum height of shut off the engine $\text{ =1}\text{.5}\text{m}$ (two significant figures)

Part (c)

To determine

Maximum height of lunar module to shut off the engine if the velocity of the lander relative to the surface at that moment is $2.0$ m/s upward.

Answer to Problem 55P

Solution: 1.5 m

Explanation of Solution

Given: Acceleration due to gravity at moon $\text{=1}\text{.62}{\text{m/s}}^{\text{2}}$

velocity of the lander relative to the surface at that moment is 2 m/s upwards.

Formula Used:

During landing, only gravitational force due moon is acting conservation of energy.

  $K.E_1+P.E_1=K.E_2+P.E_2$

Calculation:

Initial kinetic energy:

  $\begin{array}{c}\left(K.E_1\right)=\frac{1}{2}m{v}_1{}^2\\ \text{ =}\frac{\text{1}}{\text{2}}\text{×}m{\left(2\right)}^{\text{2}}\end{array}$

Initial potential energy:

  $\begin{array}{c}\left(P.E_1\right)=mg{h}_1\\ =m\text{×1}\text{.62×}{h}_1\end{array}$

Final kinetic energy:

  $\begin{array}{c}\left(K.E_2\right)=\frac{1}{2}m{v}_2{}^2\\ =\frac{1}{2}\times m\times v_2{}^2\\ =\frac{\text{1}}{\text{2}}\text{×}m\text{×}{\left(\text{3}\right)}^{\text{2}}\end{array}$

Final potential energy:

  $\begin{array}{c}\left(P.E_2\right)=mg{h}_2\\ =m\text{×1}\text{.62×}\left(\text{0}\right)\\ \text{=}\text{0}\end{array}$

Apply conservation of energy

  $K.E_1+P.E_1=K.E_2+P.E_2$

  $\begin{array}{l}\frac{1}{2}m{\left(2\right)}^2+m\text{×1}\text{.62×}{h}_1\text{=}\frac{1}{2}\times m\times {\left(\text{3}\text{.0}\right)}^{\text{2}}\text{+0}\\ 1.62\times h_1=\frac{1}{2}\left({\left(3.0\right)}^2-{\left(2.0\right)}^2\right)\\ h_1=\frac{\frac{1}{2}\left[{\left(3.0\right)}^2-{\left(2.0\right)}^2\right]}{1.62}\\ \text{ =1}\text{.5}\text{m}\end{array}$

Conclusion:Maximum height of shut off the engine $\text{ =1}\text{.5}\text{m}$ .