College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Suppose you are talking by interplanetary telephone to a friend who lives on the Moon. He tells you that he has just won a newton of gold in a contest. Excitedly, you tell him that you entered the Earth version of the same contest and also won a newton of gold! Who is richer? (a) You are. (b) Your friend is. (c) You are equally rich.
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- Consider the observation that the acceleration due to the gravitational force acting on a mass around a host planet decreases with the square of the separation between the objects. We can ask ourselves: why is it still accurate to consider a gravitational acceleration value of 9.8\frac{m}{s^2}9.8s2m for all of our projectile motion problems and all of our gravitational potential energy from prior modules? Let's analyze a situation and justify this analysis method: consider an object being launched from ground level to an altitude of 10,000 meters, roughly the cruising altitude of most jet liners, and far above our everyday experiences on Earth's surface. Compare the gravitational acceleration of the object at Earth's surface (the radius of Earth is about r_E=6.37\times10^6mrE=6.37×106m) to the acceleration value at the 10,000 meter altitude by determining the following ratio: g10,000m/gsurfacearrow_forwardWe will use differential equations to model the orbits and locations of Earth, Mars, and the spacecraft using Newton’s two laws mentioned above. Newton’s second law of motion in vector form is: F^→=ma^→ (1) where F^→ is the force vector in N (Newtons), and a^→ is the acceleration vector in m/s^2,and m is the mass in kg. Newton’s law of gravitation in vector form is: F^→=GMm/lr^→l*r^→/lr^→l where G=6.67x10^-11 m^3/s^2*kg is the universal gravitational constant, M is the mass of the larger object (the Sun), and is 2x10^30 kg, and m is the mass the smaller one (the planets or the spacecraft). The vector r^→ is the vector connecting the Sun to the orbiting objects. Step one ) The motion force in Equation(1), and the gravitational force in Equation(2) are equal. Equate the right hand sides of equations (1) and (2), and cancel the common factor on the left and right sides. Answer: f^→=ma^→ f=Gmm/lr^→l^2 a^→=Gmm/lr^→l^2 x r^→/lr^→l r^→=r^→/lr^→l * Gmm Could you please…arrow_forwardJupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×1022kg8.93×1022kg and a radius of 1821 km. How high would this material go on earth if it were ejected with the same speed as on Io? (RE = 6370 km, mE=5.96×1024kg)arrow_forward
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