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A body weighs 30 kg on the surface of the earth how much would it weigh on the surface of a planet whose mass is 1/9 the mass of earth and radius is half of that of earth?
Weight on earth = 30 kg wt = 30 x g N
g on new planet, g’ = GM’/R’2=(4/9)g
Therefor weight on the new planet will be equal to 4/9 times its weight on earth
(Whereas the mass will be the same)
ROUNAK BHATTACHARJEE asked the above question.
The fact is that linear momentum is conserved in all interactions. If you concentrate on the motion of a single body alone, that will not account for an interaction. When thrown above, according to the principle of conservation of momentum, what we can say is that “The total momentum of the ball plus the thrower just before throwing is equal to the total momentum just after throwing”
After that what happens is controlled by the force of gravity acting on it. The force acting on it can change the momentum. Here the interaction is between the ball and the earth. On account of the huge mass of earth, the motion of earth to compensate the motion of ball and to conserve the momentum cannot be detected.
However, the linear momentum is conserved in all events and interactions so far!
Rajkumar asks: “accelaration due to gravity decrease on depth of earth with the relation of g’=g(1-2h/R). but with relation of g=GM/(R*R) g increase.
When the object is on the surface or above it, the entire mass of the earth attracts it. But when it is inside, the net attraction towards CENTRE is provided only by the mass of earth below it.
ie; if it is at a depth d below the surface, then it is attracted to the centre by a sphere of radius R-d only. The forces due to the rest of the mass of earth cancel out. (On one side there is a little mass and on the other side greater amount of mass but distributed faraway and both contribute equally towards force and cancel out.)