Question

select the correct answer.

object a attracts object b with a gravitational force of 5 newtons from a given distance. if the distance between the two objects is reduced in half, what will be the changed force of attraction between them?

a. 2.5 newtons
b. 10 newtons
c. 15 newtons
d. 20 newtons
e. 25 newtons

Explanation:

Step1: Recall Gravitational Force Formula

The gravitational force between two objects is given by Newton's law of gravitation: \( F = G\frac{m_1m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1, m_2 \) are the masses of the objects, and \( r \) is the distance between them.

Step2: Analyze Force Change with Distance

Let the initial force be \( F_1 = 5 \, \text{N} \) with distance \( r_1 \). When the distance is reduced to half, \( r_2=\frac{r_1}{2} \). The new force \( F_2 \) is \( F_2 = G\frac{m_1m_2}{r_2^2}=G\frac{m_1m_2}{(\frac{r_1}{2})^2}=G\frac{m_1m_2}{\frac{r_1^2}{4}} = 4G\frac{m_1m_2}{r_1^2} \). Since \( F_1 = G\frac{m_1m_2}{r_1^2} \), we have \( F_2 = 4F_1 \).

Step3: Calculate New Force

Substitute \( F_1 = 5 \, \text{N} \) into \( F_2 = 4F_1 \): \( F_2 = 4\times5 = 20 \, \text{N} \)? Wait, no, wait. Wait, \( r_2 = \frac{r_1}{2} \), so \( r_2^2=\frac{r_1^2}{4} \), so \( F \) is inversely proportional to \( r^2 \), so if \( r \) is halved, \( F \) becomes \( 4 \) times? Wait, no, wait the initial force is \( F_1 = G\frac{m_1m_2}{r_1^2}=5 \). Then \( F_2 = G\frac{m_1m_2}{(r_1/2)^2}=G\frac{m_1m_2}{r_1^2/4}=4\times G\frac{m_1m_2}{r_1^2}=4\times5 = 20 \)? Wait, but wait, maybe I made a mistake. Wait, no, let's check again. Wait, the formula is \( F \propto \frac{1}{r^2} \). So \( \frac{F_2}{F_1}=\frac{r_1^2}{r_2^2} \). Given \( r_2 = \frac{r_1}{2} \), so \( \frac{F_2}{F_1}=\frac{r_1^2}{(r_1/2)^2}=\frac{r_1^2}{r_1^2/4}=4 \). So \( F_2 = 4\times F_1 = 4\times5 = 20 \)? Wait, but the options have D as 20. Wait, but wait, maybe I messed up. Wait, no, let's re-express. Suppose initial distance is \( r \), force \( F = 5 \). New distance \( r' = r/2 \). Then \( F' = G\frac{m1m2}{(r/2)^2} = G\frac{m1m2}{r^2/4} = 4\times G\frac{m1m2}{r^2} = 4F = 4\times5 = 20 \). So the new force is 20 newtons.

Answer:

D. 20 newtons