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[AL Phy] gravitational
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First of all, let's consider an object moving around the earth in a circular orbit (say a satellite). When the satellite is in a stable orbit of constant height above the ground: Centripetal force on the satellite = Gravitational force on the satellite i.e. 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Phy/Gravitation1.jpg where M = mass of earth, m = mass of satellite, r = radius of orbit and G = gravitational constant. Then consider that the total mechanical energy: Total energy = Gravitational Potential energy + Kinetic energy i.e. 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Phy/Gravitation2.jpg So we can see that, the total mechanical energy, is equal to the K.E. of the satellite (in value). Therefore, if we are taking the value into account only, the K.E. of the satellite will increase if r decreases, i.e. The satellite moves faster.
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[AL Phy] gravitational
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given mechanical E directly proportional to-1/r if r decreases how is KE, expain in detail最佳解答:
First of all, let's consider an object moving around the earth in a circular orbit (say a satellite). When the satellite is in a stable orbit of constant height above the ground: Centripetal force on the satellite = Gravitational force on the satellite i.e. 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Phy/Gravitation1.jpg where M = mass of earth, m = mass of satellite, r = radius of orbit and G = gravitational constant. Then consider that the total mechanical energy: Total energy = Gravitational Potential energy + Kinetic energy i.e. 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Phy/Gravitation2.jpg So we can see that, the total mechanical energy, is equal to the K.E. of the satellite (in value). Therefore, if we are taking the value into account only, the K.E. of the satellite will increase if r decreases, i.e. The satellite moves faster.
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