求英文物理题,关于行星的Pluto,the second largest dwarf planet in the Solar System,has a radius R = 1140 km and an acceleration due to gravity on its surface of magnitude g = 0.66 m/s2.a) Use these numbers to calculate the escape speed from
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求英文物理题,关于行星的Pluto,the second largest dwarf planet in the Solar System,has a radius R = 1140 km and an acceleration due to gravity on its surface of magnitude g = 0.66 m/s2.a) Use these numbers to calculate the escape speed from
求英文物理题,关于行星的
Pluto,the second largest dwarf planet in the Solar System,has a radius R = 1140 km and an acceleration due to gravity on its surface of magnitude g = 0.66 m/s2.
a) Use these numbers to calculate the escape speed from the surface of Pluto.
b) If an object is fired directly upward from the surface of Pluto with half of this escape speed,to what maximum height above the surface will the object rise?(Assume that Pluto has no atmosphere and negligible rotation.)
求英文物理题,关于行星的Pluto,the second largest dwarf planet in the Solar System,has a radius R = 1140 km and an acceleration due to gravity on its surface of magnitude g = 0.66 m/s2.a) Use these numbers to calculate the escape speed from
冥王星,太阳系中第二大的矮行星,有一个半径为1140公里和重力加速度在其表面级G = 0.66米/ S2.
b）如果一个对象是直接向上发射的逃逸速度的一半,冥王星的表面,是最大的表面之上的高度将物体上升?（假设冥王星没有大气和可以忽略不计的旋转.）
(1)逃逸速度,即万有引力不足以提供圆周运动的向心力,用重力约等于万有引力,即GmM/R^2=mg=mv^2/R
得到v=√(gR)=867.41m/s
(2)根据机械能守恒定律,上升的过程,动能转化为引力势能.
或动能定理,万有引力做功=动能变化
1/2*mv^2=∫F*dr,F=GmM/r^2,dr从R积分到R+h.h就是最大高度.
并将GmM/R^2=mg,求出GM=gR^2,带入积分式
得到2gR^2(1/R-1/(R+h))=v^2
得到h=RV^2/(2gR-v^2)=R【计算过程自己看看,也不知道我算对了没】