11Any stone will fall to the planet only if acceleration due to planet is more than moon
Therefore
a due to moon = Gm/r2
And
A due to planet =GM/x2
GM/x2>Gm/r2
M/m>x2/r2
x<r√(M/m)
x = D + R
D<r√(M/m) - R
Suppose there was a planet of mass M which was free from any other force from a planet or star...
and this planet had a moon of mass m.
Now, the rotation is such that the same surface of the moon is visible from the planet...
The radius of the planet is R, the moon is r
Find the minimum separation such that any lose stone on the moon is pulled by the planet!!
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UP 0 DOWN 0 0 10
10 Answers
11
3field due to moon on its surface is GMmoon/r2=E1
field due to planet on moon's surface is GMplanet/(D-r)2=E2
E1<E2
we will get D=Distance between the com of moon and earth.
62no sankara...
Hint: You cannot always apply ΣF =0
That is what you all are doing.. net force is zero... (Which is not true here!) why?
1net force cant be 0 here because both the planet and the moon are accelerating with respect to each other............am i right?
3sorry i didnt have patience to read the above disc!!....sorry if it has gone terribly rong!
GMm(stone)/[d-r-R]2 = Gmm(stone)/r2 + m(stone)W^2(d-r).....(1)
also
GMm/d2 = mw2d ......(2)
eleminate w frm both equations and get the disired 'd'.......