Gravity equation

1. Three particles each with a mass of 1 kg are at the vertices of an equilateral triangle whose sides are 1 m long. How large is the gravitational force experienced by each point particle (in G)?

Solution

The magnitude of the gravitational force experienced by one of the particles.

F₁₂ = G (m₁)(m₂) / r² = G (1)(1) / 1² = G/1 = G

F₁₃ = G (m₁)(m₃) / r² = G (1)(1) / 1² = G/1 = G

Resultant gravitational force at point 1:

F₁ = √1²+1² = √1+1 = √2 Newton

2. The figure below depicts three objects m₁ = 6 kg; m₂ = 3 kg and m₃ = 4 kg lie on a straight line. Determine the magnitude and direction of the resultant gravitational force experienced by m2! (state in G)

Known:

m₁ = 6 kg

m₂ = 3 kg

m₃ = 4 kg

gravitational constant = G

r₂₁ = 4 m

r₂₃ = 2 m

Wanted: F resultant gravity experienced by m₂

Solution:

The gravitational force between m₂ and m₃:

F = G (3)(4) / 2² = G 12 / 4 = 3G

The gravitational force between m₂ and m₁:

F = G (3)(6) / 4² = G 18 / 16 = 1,125G

3. Object A with a mass of 1 kg and object B with a mass of 2 kg are separated by a distance of 2 m from one another. Point P is 2 m from object A and 2 m from object B. How strong is the gravitational field at point P?

Known:

m_A = 1 kg

m_B = 2 kg

r_PA = 2 m

r_PB = 2 m

Gravity constant = G

Wanted: E gravity at point P

Solution:

E_PA = G (m_A) / r² = G (1) / 2² = G/4 = 0,25G

E_PB = G (m_B) / r² = G (2) / 2² = 2G/4 = 0,5G

Resultant gravitational field strength at point P:

E = √0,25G²+0,5G² = √0,0625G²+0,25G² = √0,3125G² = 0,56G N/kg