A particle of mass m1 is kept at x = 0 and another of mass m2 at x = d. When a third particle is kept at x = d/4, it experiences no net gravitational force due to the two particles. Find m2/m1.
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Gravitational force on m(3) due to m((1)
F = Gm(1)(3)/(d/4)²
Gravitational force on m(3) due to m(2)
F = Gm(2)m(3)/( d-d/4)²
ATQ ,
if there is no gravitational force
F = F'
or Gm(1)m(3)/(d/4)² = Gm(2)m(3)/(d-d/4)²
m1 =m2/9
m1/m2 = 1/9
F = Gm(1)(3)/(d/4)²
Gravitational force on m(3) due to m(2)
F = Gm(2)m(3)/( d-d/4)²
ATQ ,
if there is no gravitational force
F = F'
or Gm(1)m(3)/(d/4)² = Gm(2)m(3)/(d-d/4)²
m1 =m2/9
m1/m2 = 1/9
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