Calculating Gravitational Attraction: Solving for the Answer

[Jimsac]

Here is the ex. in the book. Two 3.kg textbooks are .15 apart. What is the gravitational attraction?
The answer is given 2.7*10-8
I also know the formula is F=Gm1m2/r
I just do not understand what they are doing to get the answer.

(6.67*10-11 N-m2kg2)(3.kg)(3.kg)/.15
This is the formula I am trying to use

Please help!

 

[NateTG]

It's
$$F=\frac{m_1m_2G}{r^2}$$
distance squared.

 

[Jimsac]

I know the formula I do not understand how to solve it

 

[quantumdude]

I know the formula I do not understand how to solve it

Look closer. Your formula is wrong, and Nate's is right (note the "r²" in the denominator).

 

[Jimsac]

misunderstood

What I do not understand is what to do with the formula. How do you solve this problem? THis is just an example in the book with the answer given but I do not understand what to do with the formula to solve it. Sorry

 

[NateTG]

Plug in the numbers:
$$F=\frac{m_1m_2G}{r^2} \Rightarrow \frac{3 \times 3 \times (6.67 \times 10^{-11})}{0.15^2}$$