Two objects attract each other gravitationally with a force of 2.5 10-10 N when they are 0.29 m apart.?

3 Answers

• F(g) = gravitational force = GMm / r²

  G = gravitational constant = 6.674 x 10 ̄ ¹¹ Nm²kg ̄ ²

  => 2.5 x 10 ̄ ¹⁰ N = 6.674 x 10 ̄ ¹¹[Mm] / (0.29)²

  => Mm = 0.31503 = (m)(4 – m)

  => -m² + 4m – 0.31503 = 0

  or m² – 4m + 0.31503 = 0

  solving => M = 3.92 kg and m = 0.08 kg (or vice versa)

  hope this helps

• Gravitational force Fg between two masses M₁ and M₂ separated by a distance D is given by:

  Fg = G / D²

  where G is the universal gravitational constant {6.67 * 10^-11 N.m²/kg²}

  re-arrange:

  M₁ M₂ = Fg D² / G = ( 2.5 10^-10 0.29² / 6.67 * 10^-11 )

  M₁ * M₂ = 0.315

  M₁ = 0.315 / M₂

  Given M₁ + M₂ = 4.0

  0.315 / M₂ + M₂ = 4.0 {multiply everything by M₂}

  0.315 + M₂² = 4.0M₂

  M₂² – 4.0M₂ + 0.315 = 0

  Use the ‘quadratic formula’

  a = 1; b = -4.0; c = 0.315

  M₂ = [–b ±√(b²–4ac)] / 2a

  M₂ = 3.92 kg or 0.08 kg

  That’s actually giving you both masses, since

  If M₂ = 3.92,

  M₁ = (0.315 / 3.92) = 0.08

  If M₂ = 0.08,

  M₁ = (0.315 / 0.08) = 3.92

• F = m1m2G/d^2

  2.510^-10 = m(4-m)6.6710^-11/0.29^2

  2.50.29^2 = (4m-m^2)0.667

  0.210 -2.668m+0.667m^2 = 0

  m = (2.668±√2.668^2-0.2140.667)/1.33 = 3.92 ; 0.08 kg

  or

  m1m2 = Fd^2/G = 2.510^-10/6.6710^-11*0.29^2 = 0.315

  m1+m2 = 4

  m1 = 4-m2

  4m2-m2^2 = 0.315

  m2 = (4±√4^2-1.26)/2 = 3.920 ; 0.080 kg