1) How much potential energy is required to lift the 9000-kg Soyuz vehicle from Earth's surface to the heightof the ISS, 400 km above the surface?ME=5.69x1024 kg RE=6.37x106 m

(This negative sign signifies that we need to do work.)Hence the magnitude of the potential energy…

Answered: 1) How much potential energy is…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter6: Energy Of A System

Section: Chapter Questions

Problem 50P

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A system consists of five particles. How many terms appear in the expression for the total gravitational potential energy of the system? (a) 4 (b) 5 (c) 10 (d) 20 (e) 25
A particle of mass 2.0 kg moves under the influence of the force F(x)=(3/x)N. If its speed at x=2.0 m is v=6.0 m/s, what is its speed at x = 7.0 m?
Rank the following quantities of energy from largest to smallest. State if any are equal. (a) the absolute value of the average potential energy of the SunEarth system (b) the average kinetic energy of the Earth in its orbital motion relative to the Sun (c) the absolute value of the total energy of the SunEarth system
In each situation shown in Figure P8.12, a ball moves from point A to point B. Use the following data to find the change in the gravitational potential energy in each case. You can assume that the radius of the ball is negligible. a. h = 1.35 m, = 25, and m = 0.65 kg b. R = 33.5 m and m = 756 kg c. R = 33.5 m and m = 756 kg FIGURE P8.12 Problems 12, 13, and 14.
Find the speed needed to escape from the solar system starting from the surface of Earth. Assume there are no other bodies involved and do not account for the fact that Earth is moving in its orbit. [Hint: Equation 13.6 does not apply. Use Equation 13.5 and include the potential energy of both Earth and the Sun. Substituting the values for Earth’s mass and radius directly into Equation 13.6, we obtain vesc=2GMR=2(6.67 10 11Nm2/kg2)(5.96 10 24kg)(6.37 106m)=1.12104m/s That is about 11 km/s or 25,000 mph. To escape the Sun, starting from Earth’s orbit, we use R=RES=1.501011m and MSum=1.991030kg . The result is vesc=4.21104m/s or about 42 km/s. We have 12mvesc2GMmR=12m02GMm=0 Solving for the escape velocity,
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1) How much potential energy is required to lift the 9000-kg Soyuz vehicle from Earth's surface to the height of the ISS, 400 km above the surface? ME=5.69x1024 kg RE=6.37x106 m