# The Earth's Moon versus Phobos and gravity

Our Moon weighs 7.35 x 1022 kg and Mars' moon Phobos, weighs 10.6 × 1015 kg.

How much would need to be added to Phobos to give it the same gravitational "impact" on Mars, as the Moon currently has with the Earth?

Based upon Phobos having its currently falling orbit being stabilised or would the orbit need to pushed further out?.

If I actually write the book I will give credit!

• Bit curious on how you're planning to add mass to Phobos in your book, honestly. Oct 18 '19 at 13:53
• A game of pool, but with big rocks...
– Tedz
Oct 19 '19 at 1:27
• But, if Phobos needs to be 38,656.7 times more massive than it is now, then its orbit would have to be changed to be much further out as it's only 3000 miles'ish above Mars as it is.
– Tedz
Oct 19 '19 at 1:33
• and it still needs to make Mars "flex".
– Tedz
Oct 19 '19 at 1:34
• I'd look up "Rosche Limit" too, it may be a good twist. If they get beyond that limit the tidal forces would eventually turn Phobos into a ring by breaking it up. Oct 20 '19 at 14:48

Using Newton's Law of universal gravitation, we can calculate the gravitational force between those bodies. The equation is (Note: I use "," to seperate digits into threes and use "." for decimals): $$F=G{{ M_1M_2}\over{r^2}}$$ Where $$F$$ is Force of gravity measured in Newtons, $$G$$ is Newton's gravitational constant and is equal to $$6.67259\times10^{-11}$$ $$Nm^2/kg^2$$), $$M_1$$ and $$M_2$$ are the masses of the bodies (in kg) and finally $$r$$ is the radius or distance between the centres of the masses.

So the Moon on average is 384,000,000 m away from the Earth, the mass of Earth and the Moon are $$5.972\times10^{24}$$ $$kg$$ and $$7.348\times10^{22}$$ $$kg$$ respectively. So let's substitute the values into the given equation.

Force of gravity between Earth and Moon: $$F = 6.67259\times10^{-11} {(5.972\times10^{24}){(7.348\times10^{22})} \over {384,000,000^2}}$$

and therefore, $$F$$ is equal to $$1.98573\times10^{20}$$ $$N$$.

Now onto the gravitational strength between Mars and Phobos. Mars and Phobos have a mass of $$6.39\times10^{23}$$ $$kg$$ and $$10.6\times10^{15}$$ $$kg$$ respectively. Phobos is 9380 km far from Mars. So let's substitute the values into the equation.

Force of gravity between Mars and Phobos: $$F=6.67259\times10^{-11}{{(6.39\times10^{23}){(10.6\times10^{15})}\over9,380,000^2}}$$

and therefore, $$F$$ is equal to $$5.13683\times10^{15}$$ $$N$$.

Now if we want to solve for $$M_2$$ (which is Phobos' mass) we need to rearrange the equation to solve for $$M_2$$ and substitute $$F$$ from equation 1 to equation 2 which was $$1.98573\times10^{20}$$ $$N$$.

So the new rearranged equation would be: $$M_2 = {{Fr^2}\over{GM_1}}$$ so let's substitute the values. $$M_2 = {{(1.98573\times10^{20})(9,380,000^2)\over{(6.67259\times10^{-11}){(6.39\times10^{23})}}}}$$ and therefore, $$M_2 = 4.09761\times10^{20}kg$$

Meaning if Phobos were at the same distance as it is now, it would need a mass of $$4.09761\times10^{20}$$ $$kg$$ to have the same gravitational force between it and Mars as between the Moon and the Earth. In other words, Phobos needs to be 38,656.7 times more massive than it is in reality.

• You could throw density into the mix too for some added fun. 1.88 g/cm^3 is Phobos' density (the moon is roughly double that at 3.34 g/cm^3). The radius of the moon is also 156.43 times that of Phobos. So if you wanted to keep Phobos at roughly the same size (7mi radius) you'd need an element with a density of 83.2 g/cm^3 (or ~4x the density of the densest substance known, Osmium @ 22.5g/cm^3). In other words, Phobos is going to get large enough to the point that it probably won't maintain that same density and compact itself (as it's currently a glorified ball of loose rubble). Oct 18 '19 at 13:51