Solve for the first case (before the separation doubled):

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F_{1} &= G ~ dfrac{ m_{1} ~ m_{2} }{ r_{1}^{2} } \
end{align*}
$$

Solve for the second case (after the separation doubled):

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F_{2} &= G ~ dfrac{ m_{1} ~ m_{2} }{ r_{2}^{2} } \
&= G ~ dfrac{ m_{1} ~ m_{2} }{ left( 2 r_{1} right)^{2} } \
&= G ~ dfrac{ m_{1} ~ m_{2} }{ 4 r_{1}^{2} } \
&= dfrac{1}{4} ~ G ~ dfrac{ m_{1} ~ m_{2} }{ r_{1}^{2} } \
&= dfrac{1}{4} ~ F_{1}
end{align*}
$$

So, if the separation between the two masses is doubled, then the gravitational force between them will decrease by the factor of 4.

Solve for the two gravitational forces act on the same object:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
end{align*}
$$

Let the given object is placed on a point $A$. According to the superposition principle, the total gravitational force that acts on the object is given by

$$
begin{align*}
F_{A} &= F_{A,B} + F_{A,C} \
&= G ~ dfrac{ m_{A} ~ m_{B} }{ r_{A,B}^{2} } +G ~ dfrac{ m_{A} ~ m_{C} }{ r_{A,C}^{2} } \
end{align*}
$$

So, the total gravitational force that acts on the given object is depending on the direction of each gravitational force, the mass of each acting object, and the separation between the acting object and the given object. In general, we use the concept of superposition and the vectors addition.

Solve for the two gravitational forces act on the same object:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
end{align*}
$$

So, the gravity acts on the two objects by an attraction force to attract them to each other in the direction of the center of their masses.

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
end{align*}
$$

So, the gravitational force that acts on the given object is depending on the direction, the mass of each acting object, and the separation between the acting object and the given object. In general, The universal gravitational force is directly proportional to the mass of the two objects ( if the mass increases, then the gravitational force increasing) and inversely proportional to the separation between the two objects (if the distance increase, then the gravitational force decreasing).

According to Newton’s second law, the force due to the gravity (Weight) is given by

$$
begin{align*}
F &= W \
&= m_{1} ~ a \
end{align*}
$$

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{E} }{ r^{2} } \
&= m_{1} ~ a \
&= G ~ dfrac{ m_{1} ~ m_{E} }{ left( R_{E} + h right)^{2} } \
end{align*}
$$

As we know the gravitational force is inversely proportional to the separation between the two objects. Since the height of the spacecraft on Earth is equal to zero, then the gravitational force that acts on the spacecraft is less than the gravitational force that acts on the spacecraft on the Earth. If the spacecraft moves very far away from the Earth, then the separation between them approximately equals infinity and the gravitational force between them is equal to zero. So, the astronauts are weightless. Since the orbit of the spacecraft, motion is near to the Earth, then the weight of the astronauts is less than their weight on the Earth.

Solve for the first object that is placed in $B$ point:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
end{align*}
$$

According to the superposition principle, the total gravitational force that acts on the object that is placed in $B$ point is given by

$$
begin{align*}
F_{B} &= F_{B,A} + F_{B,C} \
&= – G ~ dfrac{ m_{A} ~ m_{B} }{ r_{A,B}^{2} } +G ~ dfrac{ m_{A} ~ m_{C} }{ r_{B,C}^{2} } \
&= G left(- dfrac{ m_{A} ~ m_{B} }{ r_{A,B}^{2} } + dfrac{ m_{B} ~ m_{C} }{ r_{B,C}^{2} } right) \
&= G left( dfrac{- m ~ 2m }{ r^{2} } + dfrac{ 2m ~ 2m }{ r^{2} } right) \
&= G ~ dfrac{ 2m^{2} }{ r^{2} } ~ left(- 1 +2 right) \
&= G ~ dfrac{ 2m^{2} }{ r^{2} }
end{align*}
$$

So, the total gravitational force that acts on the object that is placed in $B$ point is $G ~ dfrac{ 2m^{2} }{ r^{2} }$ directed to the left.

Solve for the first object that is placed in $A$ point:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
end{align*}
$$

According to the superposition principle, the total gravitational force that acts on the object that is placed in $A$ point is given by

$$
begin{align*}
F_{A} &= F_{A,B} + F_{A,C} \
&= G ~ dfrac{ m_{A} ~ m_{B} }{ r_{A,B}^{2} } +G ~ dfrac{ m_{A} ~ m_{C} }{ r_{A,C}^{2} } \
&= G left( dfrac{ m_{A} ~ m_{B} }{ r_{A,B}^{2} } + dfrac{ m_{A} ~ m_{C} }{ r_{A,C}^{2} } right) \
&= G left( dfrac{ m ~ 2m }{ r^{2} } + dfrac{ m ~ 2m }{ left( 2 r right)^{2} } right) \
&= G ~ dfrac{ 2m^{2} }{ r^{2} } ~ left( 1 + dfrac{1}{ left( 2 right)^{2} } right) \
&= G ~ dfrac{ 2m^{2} }{ r^{2} } ~ dfrac{5}{4} \
&= G ~ dfrac{ 5 ~ m^{2} }{ 2 ~ r^{2} }
end{align*}
$$

So, the total gravitational force that acts on the object that is placed in $A$ point is $G ~ dfrac{ 5 ~ m^{2} }{ 2 ~ r^{2} }$.

Solve for the first object that is placed in $B$ point:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
end{align*}
$$

According to the superposition principle, the total gravitational force that acts on the object that is placed in $B$ point is given by

$$
begin{align*}
F_{B} &= F_{B,A} + F_{B,C} \
&= – G ~ dfrac{ m_{A} ~ m_{B} }{ r_{A,B}^{2} } +G ~ dfrac{ m_{A} ~ m_{C} }{ r_{B,C}^{2} } \
&= G left(- dfrac{ m_{A} ~ m_{B} }{ r_{A,B}^{2} } + dfrac{ m_{B} ~ m_{C} }{ r_{B,C}^{2} } right) \
&= G left( dfrac{- m ~ 2m }{ r^{2} } + dfrac{ 2m ~ 2m }{ r^{2} } right) \
&= G ~ dfrac{ 2m^{2} }{ r^{2} } ~ left(- 1 +2 right) \
&= G ~ dfrac{ 2m^{2} }{ r^{2} }
end{align*}
$$

So, the total gravitational force that acts on the object that is placed in $B$ point is $G ~ dfrac{ 2m^{2} }{ r^{2} }$.

Solve for the first object that is placed in $C$ point:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
end{align*}
$$

According to the superposition principle, the total gravitational force that acts on the object that is placed in $C$ point is given by

$$
begin{align*}
F_{C} &= F_{C,A} + F_{C,B} \
&= – G ~ dfrac{ m_{A} ~ m_{C} }{ r_{C,A}^{2} } -G ~ dfrac{ m_{B} ~ m_{C} }{ r_{B,C}^{2} } \
&= G left(- dfrac{ m_{A} ~ m_{C} }{ r_{A,C}^{2} } – dfrac{ m_{B} ~ m_{C} }{ r_{B,C}^{2} } right) \
&= G left( dfrac{ -m ~ 2m }{ left( 2 r right)^{2} } – dfrac{2 m ~ 2m }{ r^{2} } right) \
&= G ~ dfrac{ 2m^{2} }{ r^{2} } ~ left(- dfrac{1}{ left( 2 right)^{2} } -2 right) \
&= G ~ dfrac{ 2m^{2} }{ r^{2} } ~ dfrac{-9}{4} \
&= – G ~ dfrac{ 9 ~ m^{2} }{ 2 ~ r^{2} }
end{align*}
$$

So, the total gravitational force that acts on the object that is placed in $C$ point is $- G ~ dfrac{ 9 ~ m^{2} }{ 2 ~ r^{2} }$.

Since $G ~ dfrac{ 5 ~ m^{2} }{ 2 ~ r^{2} } > G ~ dfrac{ 2m^{2} }{ r^{2} } > – G ~ dfrac{ 9 ~ m^{2} }{ 2 ~ r^{2} }$, Then $F_{A} > F_{B} > F_{C}$.

The mass of the can of the soup is $m_{1} = 0.18 mathrm{~kg}$. The mass of the jar of the pickles is $m_{2} = 0.34 mathrm{~kg}$. The distance between the can and the jar is $r = 0.42 mathrm{~m}$.

$textbf{Required: }$

(a) Finding the magnitude gravitational force that the can exerts on the jar.

(b) Finding the magnitude gravitational force that the jar exerts on the can.

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
&= 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} ~ dfrac{ 0.18 mathrm{~kg} times 0.34 mathrm{~kg} }{ left( 0.42 mathrm{~m} right)^{2} } \
&= 0.028 mathrm{~N}
end{align*}
$$

So, the magnitude of the gravitational force by the can on the jar is $0.028 mathrm{~N}$.

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
&= 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} ~ dfrac{ 0.18 mathrm{~kg} times 0.34 mathrm{~kg} }{ left( 0.42 mathrm{~m} right)^{2} } \
&= 0.028 mathrm{~N}
end{align*}
$$

So, the magnitude of the gravitational force on the can by the jar is $0.028 mathrm{~N}$.

(b) the magnitude of the gravitational force on the can by the jar is $0.028 mathrm{~N}$.

The mass of each flower pot is $m_{1} = m_{2} = 1.6 mathrm{~kg}$.

$textbf{Required: }$

(a) Finding the magnitude gravitational force between the two flower pots, if the distance between them is $r= 1 mathrm{~m}$.

(b) Finding the magnitude gravitational force between the two flower pots, if the distance between them is $r= 3 mathrm{~m}$.

Solve for the distance between them is $r= 1 mathrm{~m}$:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
&= 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} ~ dfrac{ 1.6 mathrm{~kg} times 1.6 mathrm{~kg} }{ left( 1 mathrm{~m} right)^{2} } \
&= 1.708 times 10^{-10} mathrm{~N}
end{align*}
$$

So, the magnitude of the gravitational force between the two flower pots is $1.708 times 10^{-10} mathrm{~N}$.

Solve for the distance between them is $r= 3 mathrm{~m}$:

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
&= 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} ~ dfrac{ 1.6 mathrm{~kg} times 1.6 mathrm{~kg} }{ left( 3 mathrm{~m} right)^{2} } \
&= 1.897 times 10^{-11} mathrm{~N}
end{align*}
$$

So, the magnitude of the gravitational force between the two flower pots is $1.897 times 10^{-11} mathrm{~N}$.

(b) The magnitude of the gravitational force between the two flower pots is $1.897 times 10^{-11} mathrm{~N}$.

The mass of the first player is $m_{1} = 66 mathrm{~kg}$. The mass of the second player is $m_{2} = 72 mathrm{~kg}$. The gravitational force between the two player is $F = 3.3 times 10^{-7} mathrm{~N}$.

$textbf{Required: }$

Finding the distance between the between the two players.

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} }
end{align*}
$$

Rearrange and solve for the distance between the between the two players:

$$
begin{align*}
r &= sqrt{ G ~ dfrac{ m_{1} ~ m_{2} }{ F} } \
&= sqrt{ 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} ~ dfrac{ 66 mathrm{~kg} times 72 mathrm{~kg} }{ 3.3 times 10^{-7} mathrm{~N} } } \
&= 0.972 mathrm{~m}
end{align*}
$$

So, the distance between the two players is $0.98 mathrm{~m}$.

The mass of the baseball is $m_{1} = 0.15 mathrm{~kg}$. The distance between the batter and the ball is $r = 0.77 mathrm{~m}$. The gravitational force on the batter and by the ball is $F = 1.1 times 10^{-9} mathrm{~N}$.

$textbf{Required: }$

Finding the mass of the batter.

As Newton’s law of universal gravitation, The force of gravity between two objects is given by

$$
begin{align*}
F &= G ~ dfrac{ m_{1} ~ m_{2} }{ r^{2} } \
end{align*}
$$

Rearrange and solve for the mass of batter:

$$
begin{align*}
m &= dfrac{ F ~ r^{2} }{ m_{1} ~ G} \
&= dfrac{ 1.1 times 10^{-9} mathrm{~N} times left( 0.77 mathrm{~m} right)^{2} }{ 0.15 mathrm{~kg} times 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} } \
&= 65.186 mathrm{~kg}
end{align*}
$$

So, the mass of the batter is $65.186 mathrm{~kg}$.