# Moon's mass and tides

#### Lmoy

##### Donator
Donator
If the moon were say, 1/10 as massive as it is, how far exactly would it have to be from the Earth to have the same tidal effect? Would its density have a significant effect too? And what formula could be used to calculate all this for different masses?

#### Urwumpe

##### Not funny anymore
Donator
If the moon were say, 1/10 as massive as it is, how far exactly would it have to be from the Earth to have the same tidal effect? Would its density have a significant effect too? And what formula could be used to calculate all this for different masses?

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...

$F = \frac{G M m}{r^2}$

And of course because of that:

$\frac{1}{{r_l}^2} = \frac{1}{10 {r_2}^2} \Leftrightarrow {r_l}^2 = 10 {r_2}^2 \Leftrightarrow \frac{r_l}{\sqrt{10}} = {r_2}$

Last edited:

#### IjonTichy

##### New member
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...

$F = \frac{G M m}{r^2}$

And of course because of that:

$\frac{1}{{r_l}^2} = \frac{1}{10 {r_2}^2} \Leftrightarrow {r_l}^2 = 10 {r_2}^2 \Leftrightarrow \frac{r_l}{\sqrt{10}} = {r_2}$

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 $a_t$ caused by an object with mass $M$ at distance $r$ on an object with size $d$ is

$a_t = \frac{2GMd}{r^3}$
if $d$ is small compared to $r$, i.e. the tidal force decreases with $r^3$ not with $r^2$.

#### Lmoy

##### Donator
Donator
Thanks for the responses
So what is the moon's current tidal acceleration on the Earth?

#### RisingFury

##### OBSP developer
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 $a_t$ caused by an object with mass $M$ at distance $r$ on an object with size $d$ is

$a_t = \frac{2GMd}{r^3}$
if $d$ is small compared to $r$, i.e. the tidal force decreases with $r^3$ not with $r^2$.

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.

#### Starman

##### New member
Thanks for the responses
So what is the moon's current tidal acceleration on the Earth?

The net acceleration, that of the tide on the body taken as a whole, will be zero, since tides are a manifestation of the variation of gravitational acceleration with distance. The tidal accelerations are relative to the center of the body being considered.

#### XSSA

##### New member
I got 1.10067403e-6.
What is this in?
Mass of Moon: 7.348e22
Distance between: 384403000
So did i do it wrong?

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