Question

Can someone help meeeeee... show how to solve it plzzzzzzzz

Answer 1

Right answer: 64 units

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have a gravitation force $F_{1}=16units$, given by the formula written at the beginning. Let’s rename the distance $r$ as $d$:

$F_{1}=G\frac{m_{1}m_{2}}{d^2}$     (1)

And we are asked to find the gravitation force $F_{2}$ with a given distance of $\frac{d}{2}$:

$F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}$      

$F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}$     (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate $G m_{1}m_{2}$ in both equations:

From (1):

$Gm_{1}m_{2}=F_{1}{d}^{2}$     (3)

From (2):

$Gm_{1}m_{2}=F_{2}\frac{{d}^{2}}{4}$     (4)

If (3)=(4):

$F_{1}{d}^{2}=F_{2}\frac{{d}^{2}}{4}$     (5)

Now we have to find $F_{2}$:

$F_{2}=F_{1}{d}^{2}\frac{4}{{d}^{2}}$      

$F_{2}=4F_{1}$     (6)

If $F_{1}=16 units$:

$F_{2}=(4)(16 units)$        

$F_{2}=64 units$>>>>This is the new force of attraction