Question

The sun’s mass is about 2.7 x 107 times greater than the moon’s mass. The sun is about 400 times farther from Earth than the moon. How does the gravitational force exerted on Earth by the sun compare with the gravitational force exerted on Earth by the moon?

Answer 1

@radar

Answer 2

Hello! We're looking for a comparison, so do you think that it would be okay to find a ratio? That's what my guess is.

Answer 3

yes it is

Answer 4

Alright! So, the first step is to identify what we're given, because that and some general knowledge is what we have to start with.

We're given a couple numbers, can you identify what one of them is?

Answer 5

give me a seccond im in class connect ill be out in five

Answer 6

Okay!

Answer 7

2.7 x 107 and 400

Answer 8

@theEric

Answer 9

Okay, so those are the numbers (where the first number is probably \(22\times 10^7\).

We are told that "The sun’s mass is about 2.7 x 107 times greater than the moon’s mass." So, \(sun's\ mass = 2.7\times10^7\ moon's\ mass\). So, to make the algebra easier to work with lets rename the variables.
\(m_s\equiv sun's\ mass\)
\(m_m\equiv moon's\ mass\)
So now we can write \(m_s=2.7\times10^7\ m_m\)

Answer 10

With some algebra, we can now say what \(2.7\times 10^7\) means to physics:\[\frac {m_s}{ m_m} = 2.7\times10^7\]

Answer 11

Is that okay with you?

Answer 12

perfectly fine

Answer 13

Alright! Now, it's your turn. The next number is about distance (a good variable would be \(d\) or \(r\)).
So given that "the sun is about 400 times farther from Earth than the moon," use the same process that I did (or any other, really) to find what the "400" means to us.

Answer 14

Actually, don't even go that far.

Answer 15

Fgs/Fgm = 2.7E7/400² = 169

Answer 16

If you just write that statement, you have the information that you need.

Answer 17

so was that right

Answer 18

Yep! I didn't pull out the calculator, but that looks correct!

Answer 19

Looks good. Congrats! Take care!

Answer 20

Did you have any questions?

Answer 21

could ya help me with like five more questions and thank you so much

Answer 22

Yep! Just ask them separately for the sake of other people who find each post.

And, if you have already worked stuff out, let me know so I'm not tempted to drag you through the setup again! :)

Answer 23

haha dont mind draggin me it might teach me a thing or two im all about learning seya on the next post

Answer 24

For anyone who looks up this answer after the fact, there are at least two simple algebraic ways to solve it, and I'll go through one.

From the second piece of information,
\(distance\ from\ sun = 400 \times distance\ from\ moon\)
so
\(r_s = 400\ r_m\)

That's the given. Now we use our general knowledge. We want to compare gravitational force, so let's look at the ratio \[\frac{F_{g,s}}{F_{g,m}}\]

Well, for any gravitational force between two masses,\[F_g=G\frac{m_1\ m_2}{r^2}\]That's the universal law of gravitation (so look that up if your unsure what it all means).

For the gravitational force between the sun and the Earth, one mass is \(m_s\) and the other is \(m_E\) (for Earth's mass). For the gravitational force between the moon and the Earth, one mass is \(m_m\) and the other is still \(m_E\).

So,\[F_{g,s} = G\dfrac{m_s\ m_E}{r_{s,E}^2}\]and\[\dfrac{1}{F_{g,m}} = \dfrac1G\dfrac{r_{m,E}^2}{m_m\ m_E}\]

Put those together, some variables cancel, and your left with...
\[\frac{F_{g,s}}{F_{g,m}}=\frac{m_s}{r_{s,E}^2}\frac{r_{m,E}^2}{m_m}=\left(\frac{m_s}{m_m}\right)\left(\frac{r_{m,E}}{r_{s,E}}\right)\left(\frac{r_{m,E}}{r_{s,E}}\right)\]
Each factor in parenthesis can be derived from the information that is given in the problem.