Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 375: 67

Answer

$\approx -12\%$

Work Step by Step

First, we need to find the speed of the lander when it is in the orbit 1000 km above the moon's surface. $$\sum F_r=\dfrac{GM_{\rm moon}\color{red}{\bf\not}m_{\rm lander}}{(R_{\rm moon}+h)^2}=\color{red}{\bf\not}m_{\rm lander}a_r$$ $$ \dfrac{GM_{\rm moon} }{(R_{\rm moon}+h)^{\color{red}{\bf\not}2}}= \dfrac{v^2}{\color{red}{\bf\not}(R_{\rm moon}+h)}$$ $$v=\sqrt{ \dfrac{GM_{\rm moon} }{(R_{\rm moon}+h)}}$$ Plugging the known: $$v=\sqrt{ \dfrac{(6.67\times 10^{-11})(7.36\times 10^{22}) }{(1.74\times 10^6)+(1000\times 10^3)}}$$ $$v_i=\bf 1339\;\rm m/s$$ Now we need to make the lander move in an elliptic path around the moon where the closest point to the moon is its own surface and its far point is the original orbit of the lander. We can use the conservation of angular momentum and the conservation of energy. $$L_1=L_2$$ $$\color{red}{\bf\not}mv_1r_1\color{red}{\bf\not}\sin90^\circ =\color{red}{\bf\not}mv_2r_2\color{red}{\bf\not}\sin90^\circ $$ where $v_1$ is the speed of the lander at the far point from the moon and $v_2$ is just on the moon's surface. Hence, $$ v_1(R_{\rm moon}+h)=v_2R_{\rm moon}$$ Hence, $$v_2=\dfrac{ v_1(R_{\rm moon}+h)}{R_{\rm moon}}\tag 1$$ So that $$E_i=E_f$$ $$K_i+U_{ig}=K_f+U_{fg}$$ $$\frac{1}{2}\color{red}{\bf\not}mv_1^2+\dfrac{-GM_{\rm moon}\color{red}{\bf\not}m}{(R_{\rm moon}+h)}=\frac{1}{2}\color{red}{\bf\not}mv_2^2+\dfrac{-GM_{\rm moon}\color{red}{\bf\not}m}{R_{\rm moon}}$$ Multiplying both sides by 2 and plugging from (1): $$ v_1^2+\dfrac{-2GM_{\rm moon} }{(R_{\rm moon}+h)}= \left[\dfrac{ v_1(R_{\rm moon}+h)}{R_{\rm moon}}\right]^2+\dfrac{-2GM_{\rm moon} }{R_{\rm moon}}$$ $$ v_1^2+\dfrac{-2GM_{\rm moon} }{(R_{\rm moon}+h)}= \dfrac{ v_1^2(R_{\rm moon}+h)^2}{R_{\rm moon}^2}+\dfrac{-2GM_{\rm moon} }{R_{\rm moon}}$$ $$ v_1^2- \dfrac{ v_1^2(R_{\rm moon}+h)^2}{R_{\rm moon}^2}=\dfrac{-2GM_{\rm moon} }{R_{\rm moon}}+\dfrac{ 2GM_{\rm moon} }{(R_{\rm moon}+h)}$$ $$ v_1^2\left[1- \dfrac{ (R_{\rm moon}+h)^2}{R_{\rm moon}^2}\right]=-2GM_{\rm moon} \left(\dfrac{1}{R_{\rm moon}}-\dfrac{ 1 }{(R_{\rm moon}+h)}\right)$$ $$ v_1^2\left[ \dfrac{ R_{\rm moon}^2-(R_{\rm moon}+h)^2}{R_{\rm moon}^{\color{red}{\bf\not}2}}\right]=-2GM_{\rm moon} \left( \dfrac{ \color{red}{\bf\not}R_{\rm moon}+h- \color{red}{\bf\not}R_{\rm moon}}{\color{red}{\bf\not}R_{\rm moon}(R_{\rm moon}+h)}\right)$$ $$ v_1^2\left[ \dfrac{ R_{\rm moon}^2-(R_{\rm moon}+h)^2}{R_{\rm moon} }\right]= \left( \dfrac{ -2GM_{\rm moon}h }{ R_{\rm moon}+h }\right)$$ $$ v_1 =\sqrt{ \left( \dfrac{ -2GM_{\rm moon}h }{ R_{\rm moon}+h }\right)\left[ \dfrac{R_{\rm moon} }{ R_{\rm moon}^2-(R_{\rm moon}+h)^2}\right]}$$ Plugging the known; $$ v_1 =\sqrt{ \left( \dfrac{ -2(6.67\times 10^{-11} )(7.36\times 10^{22})(1000\times 10^3) }{ (1.74\times 10^6)+(1000\times 10^3)}\right)\left[ \dfrac{(1.74\times 10^6) }{ (1.74\times 10^6)^2-((1.74\times 10^6)+(1000\times 10^3))^2}\right]}$$ $$v_1=\bf 1180\;\rm m/s$$ So the fractional change in speed is given by $${\rm change\;percentage}=\dfrac{v_1-v_i}{v_1}\times 100\%$$ $${\rm change\;percentage}=\dfrac{1180-1339}{1339}\times 100\%=\color{red}{\bf -11.9}\bf \;\%$$
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