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.

F_{12} = G (m_{1})(m_{2}) / r^{2 }= G (1)(1) / 1^{2 }= G/1 = G

F_{13} = G (m_{1})(m_{3}) / r^{2 }= G (1)(1) / 1^{2 }= G/1 = G

Resultant gravitational force at point 1:

F_{1 }= √1^{2}+1^{2} = √1+1 = √2 Newton

2. The figure below depicts three objects m_{1} = 6 kg; m_{2} = 3 kg and m_{3} = 4 kg lie on a straight line. Determine the magnitude and direction of the resultant gravitational force experienced by m2! (state in G)

__Known:__

m_{1 }= 6 kg

m_{2} = 3 kg

m_{3} = 4 kg

gravitational constant = G

r_{21} = 4 m

r_{23} = 2 m

__Wanted:__ F resultant gravity experienced by m_{2}

__Solution:__

The gravitational force between m_{2} and m_{3}:

F = G (3)(4) / 2^{2} = G 12 / 4 = 3G

The gravitational force between m_{2} and m_{1}:

F = G (3)(6) / 4^{2} = 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}) / r^{2 }= G (1) / 2^{2 }= G/4 = 0,25G

E_{PB} = G (m_{B}) / r^{2 }= G (2) / 2^{2 }= 2G/4 = 0,5G

Resultant gravitational field strength at point P:

E = √0,25G^{2}+0,5G^{2} = √0,0625G^{2}+0,25G^{2} = √0,3125G^{2} = 0,56G N/kg