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(a) Newton's Law of Gravitation states that two bodies with masses $ m_1 $ and $ m_2 $ attract each other with a force$$ F = G \frac{m_1 m_2}{r^2} $$where $ r $ is the distance between the bodies and $ G $ is the gravitational constant. If one of the bodies is fixed, find the work needed to move the other from $ r = a $ to $ r = b $.(b) Compute the work required to launch a 1000-kg satellite vertically to a height of 1000 km. You may assume that the earth's mass is $ 5.98 \times 10^{24} kg $ and is concentrated at its center. Take the radius of the earth to be $ 6.37 \times 10^6 m $ and $ G = 6.67 \times 10^{-11} N \cdot m^2/kg^2 $.

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(a) $G m_{1} m_{2}\left(\frac{1}{a}-\frac{1}{b}\right)$(b) $$8.50 \times 10^{9} \mathrm{J}$$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 4

Work

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we're given the force. So the work is simply a It'll be at Artie are and plug in the force, which is to gm went ou r squared the r So that's active G and one and two over our tons. The man is one once a month's work. Before you can say G comes, I'm one too over our So when there were eight months, man would be That's the part you had for murder. And we're gonna use this one to Catholic part B. So in this part, we have g equals two 6.667 times, 10 to the minus 11 and one. This fight on 98 times 98 10 to 24 2 is 1000 a is 6.37 times 10 to the sixth and B is 713137 times to six and plugging all the data we're gonna have Devil equals two approximately April 5 times 10 to the ninth, Jule

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