consider two solid uniform spherical objects of the same density p. one has a radius r and the other 2r. they are in outer space where the gravitational force due to other objects is negligible. if they are at rest with their surfaces touching, what is the force between the objects to to the gravitational attraction?
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Answers
Answered by
11
The force between them is
F = Gm1m2/r^2 ........(1)
for
m1 = p*vol.
m1= p*4/3πr^3
for
m2 = p*vol
m2 = p*4/3π(2r)^3
putting this in eqn (1)
F = G (p*4/3πr^3)*[p*4/3π8r^3] / r^2
on multiplying m1 and m2 we have
F = G (p*4/3πr^3)^2 *8 / r^2
F = G (p*4/3π)^2 *r^6* 8 / r^2
F = G (p*4/3π)^2 * r^4 * 8
here G, p, 4/3π & 8 are constant
so
F is directly proportional to r^4
F = Gm1m2/r^2 ........(1)
for
m1 = p*vol.
m1= p*4/3πr^3
for
m2 = p*vol
m2 = p*4/3π(2r)^3
putting this in eqn (1)
F = G (p*4/3πr^3)*[p*4/3π8r^3] / r^2
on multiplying m1 and m2 we have
F = G (p*4/3πr^3)^2 *8 / r^2
F = G (p*4/3π)^2 *r^6* 8 / r^2
F = G (p*4/3π)^2 * r^4 * 8
here G, p, 4/3π & 8 are constant
so
F is directly proportional to r^4
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Answered by
2
The force between them is
F = Gm1m2/r^2 ........(1)
for
m1 = p*vol.
m1= p*4/3πr^3
for
m2 = p*vol
m2 = p*4/3π(2r)^3
putting this in eqn (1)
F = G (p*4/3πr^3)*[p*4/3π8r^3] / r^2
on multiplying m1 and m2 we have
F = G (p*4/3πr^3)^2 *8 / r^2
F = G (p*4/3π)^2 *r^6* 8 / r^2
F = G (p*4/3π)^2 * r^4 * 8
here G, p, 4/3π & 8 are constant
so
F is directly proportional to r^4
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