Answer :
Let's solve this problem step by step.
1. Initial Information:
- Initial gravitational force (F_initial) between object A and object B is 5 newtons.
- The distance between the two objects is reduced to half of its original distance.
2. Law of Gravitation:
The gravitational force between two objects is inversely proportional to the square of the distance between them. This means:
[tex]\[ F \propto \frac{1}{d^2} \][/tex]
Where [tex]\( F \)[/tex] is the gravitational force and [tex]\( d \)[/tex] is the distance between the objects.
3. Initial Distance Concept:
Let's denote the initial distance as [tex]\( d \)[/tex]. The formula for the initial force can be written as:
[tex]\[ F_initial = \frac{G \times m_A \times m_B}{d^2} \][/tex]
Since the constants [tex]\( G \times m_A \times m_B \)[/tex] are not changing, we can focus on the distance part.
4. Changed Distance:
The new distance is half of the original distance, so the new distance [tex]\( d_{new} \)[/tex] is:
[tex]\[ d_{new} = \frac{d}{2} \][/tex]
5. Changed Force Calculation:
We substitute the new distance into the formula for the gravitational force:
[tex]\[ F_{new} = \frac{G \times m_A \times m_B}{(d_{new})^2} \][/tex]
Since [tex]\( d_{new} = \frac{d}{2} \)[/tex]:
[tex]\[ F_{new} = \frac{G \times m_A \times m_B}{\left(\frac{d}{2}\right)^2} = \frac{G \times m_A \times m_B}{\frac{d^2}{4}} = G \times m_A \times m_B \times \frac{4}{d^2} \][/tex]
Now the new force is:
[tex]\[ F_{new} = 4 \times \frac{G \times m_A \times m_B}{d^2} = 4 \times F_{initial} \][/tex]
6. Substituting the Initial Force:
We know the initial force [tex]\( F_{initial} \)[/tex] is 5 newtons:
[tex]\[ F_{new} = 4 \times 5 = 20 \, \text{newtons} \][/tex]
So, the changed force of attraction when the distance is reduced to half is 20 newtons.
The correct answer is:
O D. 20 newtons
1. Initial Information:
- Initial gravitational force (F_initial) between object A and object B is 5 newtons.
- The distance between the two objects is reduced to half of its original distance.
2. Law of Gravitation:
The gravitational force between two objects is inversely proportional to the square of the distance between them. This means:
[tex]\[ F \propto \frac{1}{d^2} \][/tex]
Where [tex]\( F \)[/tex] is the gravitational force and [tex]\( d \)[/tex] is the distance between the objects.
3. Initial Distance Concept:
Let's denote the initial distance as [tex]\( d \)[/tex]. The formula for the initial force can be written as:
[tex]\[ F_initial = \frac{G \times m_A \times m_B}{d^2} \][/tex]
Since the constants [tex]\( G \times m_A \times m_B \)[/tex] are not changing, we can focus on the distance part.
4. Changed Distance:
The new distance is half of the original distance, so the new distance [tex]\( d_{new} \)[/tex] is:
[tex]\[ d_{new} = \frac{d}{2} \][/tex]
5. Changed Force Calculation:
We substitute the new distance into the formula for the gravitational force:
[tex]\[ F_{new} = \frac{G \times m_A \times m_B}{(d_{new})^2} \][/tex]
Since [tex]\( d_{new} = \frac{d}{2} \)[/tex]:
[tex]\[ F_{new} = \frac{G \times m_A \times m_B}{\left(\frac{d}{2}\right)^2} = \frac{G \times m_A \times m_B}{\frac{d^2}{4}} = G \times m_A \times m_B \times \frac{4}{d^2} \][/tex]
Now the new force is:
[tex]\[ F_{new} = 4 \times \frac{G \times m_A \times m_B}{d^2} = 4 \times F_{initial} \][/tex]
6. Substituting the Initial Force:
We know the initial force [tex]\( F_{initial} \)[/tex] is 5 newtons:
[tex]\[ F_{new} = 4 \times 5 = 20 \, \text{newtons} \][/tex]
So, the changed force of attraction when the distance is reduced to half is 20 newtons.
The correct answer is:
O D. 20 newtons
