You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Lmoy
- Start date

- Joined
- Feb 6, 2008

- Messages
- 37,845

- Reaction score
- 2,582

- Points
- 203

- Location
- Wolfsburg

- Preferred Pronouns
- Sire

The same tidal effect... that is a tough question since the tides depend a lot on geography and the inertia of the moving water.

But generally, you can say, it would have the same gravity force acting on the water, if ...

And density would have ...

Should I do your homework now? :lol:

As formula, simple simple simple...

[math]F = \frac{G M m}{r^2}[/math]

And of course because of that:

[math]\frac{1}{{r_l}^2} = \frac{1}{10 {r_2}^2} \Leftrightarrow {r_l}^2 = 10 {r_2}^2 \Leftrightarrow \frac{r_l}{\sqrt{10}} = {r_2} [/math]

Last edited:

The same tidal effect... that is a tough question since the tides depend a lot on geography and the inertia of the moving water.

But generally, you can say, it would have the same gravity force acting on the water, if ...

And density would have ...

Should I do your homework now? :lol:

As formula, simple simple simple...

[math]F = \frac{G M m}{r^2}[/math]

And of course because of that:

[math]\frac{1}{{r_l}^2} = \frac{1}{10 {r_2}^2} \Leftrightarrow {r_l}^2 = 10 {r_2}^2 \Leftrightarrow \frac{r_l}{\sqrt{10}} = {r_2} [/math]

Sorry but that's wrong. The tides are not related to the strength of the gravity force itself but to the inhomogeinity of the gravitational field. Therefore a near object with small mass can cause stronger tides than a massive object at large distance even if the latter produces a higher gravity force. For example the gravitational force of the sun acting on the earth is about 175 times stronger than the gravitational force of the moon. But the tides caused by the moon are about 2 times larger than that of the sun.

The tidal acceleration [math]a_t[/math] caused by an object with mass [math]M[/math] at distance [math]r[/math] on an object with size [math]d[/math] is

[math]a_t = \frac{2GMd}{r^3}[/math]

if [math]d[/math] is small compared to [math]r[/math], i.e. the tidal force decreases with [math]r^3[/math]

Sorry but that's wrong.

Well, I was all ready to yell at Urwumpe, but you did that job for me :thumbup:

The tidal acceleration [math]a_t[/math] caused by an object with mass [math]M[/math] at distance [math]r[/math] on an object with size [math]d[/math] is

[math]a_t = \frac{2GMd}{r^3}[/math]

if [math]d[/math] is small compared to [math]r[/math], i.e. the tidal force decreases with [math]r^3[/math]notwith [math]r^2[/math].

Just a quick follow up...

That formula is calculated by figuring out the difference of gravitational acceleration on both sides of the body and under the assumption that d << r, to simplify the equation.

Lmoy said:So what is the moon's current tidal acceleration on the Earth?

You were given the formula. Punch in the numbers and do the math yourself.

Thanks for the responses

So what is the moon's current tidal acceleration on the Earth?

The

- Replies
- 14

- Views
- 400

- Replies
- 1

- Views
- 371

- Replies
- 17

- Views
- 1K

- Replies
- 7

- Views
- 544

- Replies
- 13

- Views
- 551