Two objects attract each other gravitationally with a force of 2.5 1010 N when they are 0.29 m apart. Their total mass is 4.0 kg. Find their individual masses.
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}
rearrange:
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(4m)6.6710^11/0.29^2
2.50.29^2 = (4mm^2)0.667
0.210 2.668m+0.667m^2 = 0
m = (2.668±√2.668^20.2140.667)/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 = 4m2
4m2m2^2 = 0.315
m2 = (4±√4^21.26)/2 = 3.920 ; 0.080 kg