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Suppose you are standing on the surface of the Earth with a baseball of mass m. How fast would you have to throw the baseball horizontally so that it never landed on earth?
v=sqrt(GMe/r^2)
Me is the mass of the earth.
Simple enough, but when you do these types of problems you know that the mass in mv^2/r is the mass of the orbiting object. Is this just a convention?
I'm asking because on a separate problem I got confused about which mass I could cancel out. Here is that problem.
A new planet is discovered, 100Res away from the Earth (Res is the orbital radius of earth around the sun). With only this information, what else can we determine about this planet?
A. Both its mass and its orbital velocity
C. Its orbital velocity only
Using mv^2/r = FGmm/r^2, you can see how calculating the orbital velocity is possible. However, ifyou cancel out the wrong m (e.g mass of earth instead of mass of the new planet) you would get answer A. The answer is C.
Is the mass in mv^2r always going to be the mass of the orbiting planet?
v=sqrt(GMe/r^2)
Me is the mass of the earth.
Simple enough, but when you do these types of problems you know that the mass in mv^2/r is the mass of the orbiting object. Is this just a convention?
I'm asking because on a separate problem I got confused about which mass I could cancel out. Here is that problem.
A new planet is discovered, 100Res away from the Earth (Res is the orbital radius of earth around the sun). With only this information, what else can we determine about this planet?
A. Both its mass and its orbital velocity
C. Its orbital velocity only
Using mv^2/r = FGmm/r^2, you can see how calculating the orbital velocity is possible. However, ifyou cancel out the wrong m (e.g mass of earth instead of mass of the new planet) you would get answer A. The answer is C.
Is the mass in mv^2r always going to be the mass of the orbiting planet?
