宇宙飞船在330km空中绕地球飞,每91.1分转一圈,求向心加速度（答案含g）,下面是原题Suppose the space shuttle is in orbit 330 km from the Earth's surface,and circles the Earth about once every 91.1 minutes.Find the centripeta
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宇宙飞船在330km空中绕地球飞,每91.1分转一圈,求向心加速度（答案含g）,下面是原题Suppose the space shuttle is in orbit 330 km from the Earth's surface,and circles the Earth about once every 91.1 minutes.Find the centripeta
宇宙飞船在330km空中绕地球飞,每91.1分转一圈,求向心加速度（答案含g）,下面是原题
Suppose the space shuttle is in orbit 330 km from the Earth's surface,and circles the Earth about once every 91.1 minutes.Find the centripetal acceleration of the space shuttle in its orbit.Express your answer in terms of g,the gravitational acceleration at the Earth's surface.
宇宙飞船在330km空中绕地球飞,每91.1分转一圈,求向心加速度（答案含g）,下面是原题Suppose the space shuttle is in orbit 330 km from the Earth's surface,and circles the Earth about once every 91.1 minutes.Find the centripeta
Suppose an object（with mass m） is just above the earth‘ surface,we can get
GMm/R*2 =mg M is the earth’ mass R is the radius of the earth
we get gR*2=GM
Then we consider about the shuttle GMm1/（R+h）*2=m1（R+h）4π*2/T*2
you can get R+h
GMm1/(R+h)*2=m1a a=gR*2/(R+h)*2 fill the number into the equation,then it is solved
GRE！
地球半径为R约为6370km，飞船离地高度为h，飞船收到地球的万有引力提供向心力，产生向心加速度为a，结合黄金代换GM=gR^2，由牛顿第二定律得：GMm/(R+h)^2=ma，
得a=GM/(R+h)^2=gR^2/(R+h)^2=g*（6370/6700）^2=0,90g