Question

Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation (drawn) where F is the magnitude of the gravitational attraction on either body,m1 and m2 are the masses of the bodies, r is the distance between them, and G is the gravitational constant. In SI units, the units or force are kg m/s2, the units of mass are kg, and the units of distance are m. For this equation to have consistent units, the units of G

Answer 1

\[F= \frac{ Gm _{1}m _{2} }{ r ^{2} }\]

Answer 2

You could find it on the internet, but it would be better practice to find it yourself!
You can solve for G, and then substitute in the units!

Answer 3

\[G=\frac{Fr^2}{m_1m_2}\]

Answer 4

So, do you see what you would get by putting in the units? \[\frac{(kg\times m / s^2)(m)}{(kg)(kg)}\]
Then solve.

Answer 5

Any questions?

Answer 6

yeah, how do you know which units go where?

Answer 7

or how do you know which letters represent which letters...

Answer 8

Well, I just substituted them in. For force, you know the unit is \(kg\times m /s^2\).

Answer 9

no i have not learned that, i was only given the quantities of distance, area, volume, velocity, acceleration, and energy

Answer 10

It's much simpler to look at something like \(\large v=\frac{d}{t}\). Then \(v\) is in units of distance of units of time. Say distance was in meters, and time in seconds. It is meters over seconds, by substitution. \(\large\frac{meters}{seconds}=\frac{m}{s}=\normalsize m/s\)

Answer 11

When you say \(r^2\), you are then talking about \((meters)^2\text{ or }m^2\).

Answer 12

Does that clear up confusion? Or is it a point of knowing what the variables are?

Answer 13

no sure that is the equation that i was given

Answer 14

like how do you know F= kg×m/s2, what does r^2 represent, and m1 and m2

Answer 15

The equation looks different because I used algebra to rearrange it.
Would you like to go through that together?

I know what the variables because they are standard in physics. Are you being taught this in physics? Or math?

Answer 16

physics

Answer 17

oh... you just had to read the problem to figure out how to replace them, got it... now how do you find the answer.... like the final units?

Answer 18

\[\frac{ m ^{3} }{ kg*s ^{2} }\]

Answer 19

I gotcha. Have you learned about force, mass, and radius? The distance part is a point of confusion! People commonly use \(r\) for radius, or distance when direction of the distance isn't important.

Answer 20

is that the answer?

Answer 21

Yep! You got it! And I made a typo earlier. It was \(\large\frac{(kg\times m / s^2)(m)}{(kg)(kg)}
\normalsize \text { but should be }\large\frac{(kg\times m / s^2)(m)^2}{(kg)(kg)}
\)
But I guess you figured that out on your own!

Congrats!

Answer 22

haha thank you, i am highly surprised, i have never taken physics, it is the second day.... yay! lol

Answer 23

and thanks for your help

Answer 24

Lots of times you'll see it expressed using newtons. To see that, note that \((kg\times m/s^2) = 1N\).
That simplifies to \(\Large\frac{N\times m^2}{kg^2}=(N)\left(\frac{m}{kg}\right)^2\).

Congrats! You aced it! You're welcome, and good luck with the rest of the course!

Answer 25

thanks :)

Answer 26

:)