# History of investigation of lunar and Martian surface gravity

I wonder when it was discovered that the Moon has about a sixth of the Earth's surface gravity and that Mars has 0.38 g. I think that until Newton, everyone assumed there was the same gravity on every celestial body, the same as the Earth's. But did Newton already calculate the Moon's gravity to be about a sixth that of Earth?

Did the first flyby and impactor probes on the Moon (like Lunnik 1, 2, 3 and Ranger 7) take into account the different surface gravity around 0.166 g? Same question for Mars. I suppose they must have known the Moon's and Mars' gravity more precisely and taken them into account, otherwise it would have been too risky to send the probes there.

When was the Moon's and Mars' surface gravity calculated first and when were they measured first, and how precise were the first calculations and measurements?

• This is really several questions that ought to be split into separate posts. Jun 21, 2020 at 10:33
• @DrSheldon Sorry. I think it would look a bit weird to ask multiple similar questions. I think you need to ask them in a single one because one must elaborate on the history of determining the gravity on Moon and Mars. Your answer is good, now I'd like to learn when the Moon's (and Mars') gravity was first determined more correctly. Jun 21, 2020 at 10:37
• I think you could reasonably split it into two questions: the first calculations, and the first measurements. My answer addresses the former. Uwe's answer addresses the latter. Jun 21, 2020 at 10:44

There are two small moons of Mars called Phobos and Deimos discovered in 1877. Using orbital mechanics the mass of Mars could be calculated from the observed orbital period and the orbit diameter of the marsian moons. Using the diameter of Mars from astronomical measurements the surface gravity could be calculated.

Before the space age, the orbit diameter of the Marsian moons as well as the diameter of Mars could be estimated using telescopes from Earth only. The precision of those measurements was limited by the huge distance between Earth and Mars.

The orbital period of a small moon is:

$$T = 2\pi \sqrt{\frac{r^3}{GM}} \\ \\ r \text{ orbital radius}, M \text{ mass of planet}, G \text{ gravitational constant}$$

The mass of the moon should be neglible small compared to the mass of the planet.

Solving the equation for planets mass:

$$M = \frac{4\pi^2 r^3}{T^2G}$$

This equation is very sensitive to errors of the orbital radius due to the third power and sensitve to errors of the orbital period due to second power. Errors of the gravitational constant are less important.

The surface gravity of a celestial body is:

$$g = G \frac{M}{r^2} \\ M \text{ body mass, } r \text{ body radius}$$

A small Python script to calculate Marsian mass and gravity from the moons orbital data:

import numpy as np

pi = np.pi
G = 6.6743015E-11       # gravitational constant

def mass(radius, period):   # calculate planet mass using moon orbit data
return m

def gravity(mass, radius):  # calculate planet surface gravity from mass and radius
return g

def days_to_seconds(days):  # calculate time in seconds from days
return days * 24.0 * 3600.0

#orbital data for Deimos
r_Deimos = 23459E3  # orbital radius in meters
P_Deimos = days_to_seconds(1.2624)   # orbital period in seconds

#orbital data for Phobos
r_Phobos = 9378E3  # orbital radius in meters
P_Phobos = days_to_seconds(0.3189)   # orbital period in seconds

m_D = mass(r_Deimos, P_Deimos)
m_P = mass(r_Phobos, P_Phobos)

print('mass of Mars using the orbit of Deimos in kg',"{:1.3e}".format(m_D))
print('mass of Mars using the orbit of Phobos in kg', "{:1.3e}".format(m_P))

r_M_eq = 0.5*6792.4E3 # equatorial radius of Mars in meters

print()
print('gravity of Mars using the orbit of Deimos', "{:1.3f}".format(gravity(m_D, r_M_eq)))
print('gravity of Mars using the orbit of Phobos', "{:1.3f}".format(gravity(m_P, r_M_eq)))


results:

mass of Mars using the orbit of Deimos in kg 6.419e+23
mass of Mars using the orbit of Phobos in kg 6.426e+23

gravity of Mars using the orbit of Deimos 3.714
gravity of Mars using the orbit of Phobos 3.718


This technique was used by Dutch astronomers in 1927 to estimate the mass of Mars to an accuracy of 0.2%. This remarkable result requires an accuracy of only 0.067 % for the orbital radius. For the radius of 23459 km it is only +- 15 km.

There is no natural satellite of our Moon, so the method used above could not be used to determine the Moons gravity. But the Moon does not rotate around the center of Earth, both rotate around their common center of gravity. This movement of Earth may be measured and allows an estimation of the ratio of the mass of the Moon to the mass of Earth.

The mass of Earth could be calculated from surface gravity measurements and the radius of Earth. Using the ratio from above the Mass of Moon could be calculated.

More details in weighing the Moon and Measuring the Moon's Mass.

In 1940 a very accurate value was published, only small improvements between 1960 and 2000. Figure from Measuring the Moon's Mass.

So gravity of Moon and Mars could be estimated decades before the space age but with very limited precison.

From Wikipedia's Pierre-Simon Laplace:

Laplace's tidal equations

In 1776, Laplace formulated a single set of linear partial differential equations, for tidal flow described as a barotropic two-dimensional sheet flow. Coriolis effects are introduced as well as lateral forcing by gravity. Laplace obtained these equations by simplifying the fluid dynamic equations. But they can also be derived from energy integrals via Lagrange's equation.

For a fluid sheet of average thickness D, the vertical tidal elevation ζ, as well as the horizontal velocity components u and v (in the latitude φ and longitude λ directions, respectively) satisfy Laplace's tidal equations46...

46The Laplace Tidal Equations andAtmospheric Tides by David A. Randall.

• @LoveForChrist The Mars_Orbiter_Laser_Altimeter should deliver very precise orbital data using the laser altimeter. Using this data better values of surface gravity may be estimated.
– Uwe
Jun 22, 2020 at 14:29
• @SteveLinton Modern superconducting gravimeters are able to measure variations caused by the position of the Moon. See Wikipedia. Roughly about +- 1 µm per second square.
– Uwe
Jun 22, 2020 at 15:43
• @AntonHengst "William Shanks (25 January 1812 – June 1882)[1] was a British amateur mathematician. Shanks is famous for his calculation of π to 707 places, accomplished in 1873, which, however, was only correct up to the first 527 places.[2] This error was highlighted in 1944 by D. F. Ferguson (using a mechanical desk calculator)" So there were much more digits of Pi available than neccessary. 6 digits of Pi are more than needed,
– Uwe
Jun 22, 2020 at 17:58
• @uhoh Pierre Simon Laplace found better results for the ratio of the masses of Moon and Earth using the ocean tides. Using an extended set of tidal data in 1825, Laplace obtained the ratio 75, our present value is 81.300588, see this page.
– Uwe
Jun 22, 2020 at 21:42
• @Uwe that's really amazing! I think it's so important that I've added a bit about it to the end of your answer. Thank you for pointing it out and educating me :-)
– uhoh
Jun 22, 2020 at 22:35

But did Newton already calculate the Moon's gravity to be about a sixth that of Earth?

His calculations were off. Principia Mathematica, volume 3, proposition XXXVII is "To find the force of the Moon to move the sea". Corollary 1 finds the ratio of the Sun's force on the tides to the Moon's force on the tides. Corollary 2 finds the ratio of the Moon's force on the tides to Earth's gravity. Corollary 3 finds the ratio of the densities of the Moon and the Earth:

therefore the density of the Earth is to the density of the Moon as 4891 to 4000, or as 11 to 9. Therefore the body of the Moon is more dense and earthly, than the Earth it self.

He makes a similar claim earlier (proposition VII corollary 3: "The Moon is denser than the Earth, as shall appear afterwards."). But he was wrong; the average density of the Moon is 3346 kg/m$$^3$$ and of the Earth 5515 kg/m$$^3$$.

Corollary 4 finds the ratio of the mass of the Moon to the mass of the Earth:

the mass of matter in the Moon will be to the matter of the mass of the Earth as 1 to 39,788.

Again, he is wrong. The actual mass ratio is about 81.3.

Finally, in corollary 5, he compared the surface acceleration due to gravity:

And the accelerative gravity on the surface of the Moon will be about three times less than the accelerative gravity on the surface of the Earth.

The actual value is about 1/6, not 1/3. However, considering that all of Newton's calculations derived from observations using telescopes of his day, it's not too bad of an approximation.

According to an article in the NASA ADS journal, Measuring The Moon's Mass:

It can be seen that the lunar mass was known to about +/- 50% between 1687 and 1755, +/- 10% between 1755 and 1830, +/- 3% between 1830 and 1900, +/- 0.15% between 1900 and 1968, and +/- 0.0001% between 1968 and the present [2002].

In 1900, the size and distance of the moon was determined to high accuracy, which combined with the mass gave a good estimate of lunar surface gravity.

• The estimation of 1687 seems to be that by Isaac Newton in Principia Mathematica.
– Uwe
Jun 22, 2020 at 17:01

This answer is really a long comment: it doesn't answer the question as such, but rather tells you some ways you probably can't measure the mass of the Moon in particular, but with an interesting corollary.

One tempting thing is to say: well, the Moon's gravity should alter $$g$$ on Earth: can we measure that? Well, let's assume that the Moon and the Earth are spherically-symmetric (this is going to be a big problem) and see what the variation is. Well:

\begin{align} g_0 &= \frac{GM}{R^2}&&\text{M,R mass, radius of Earth resp.}\\ g_{+} &= \frac{GM}{R^2} + \frac{Gm}{(r + R)^2}&&\text{m,r mass, distance of Moon resp.}\\ g_{-} &= \frac{GM}{R^2} - \frac{Gm}{(r - R)^2} \end{align}

And then the interesting thing to calculate is

$$\frac{\Delta g}{g} \equiv \frac{g_{+} - g_{-}}{g_0}$$

Which tells you how accurately you need to measure $$g$$ to be able to detect the Moon.

Well, if you do this calculation you get $$\Delta g / g \approx 6.7\times 10^{-6}$$: you'd have to be able to measure $$g$$ to a few parts per million to be able to see the Moon's influence at all, and much better than that to get any reasonable estimate for the mass of the Moon.

And I made a big, wrong, assumption above: that the Earth and the Moon are spherical. Well, the Earth isn't, and it's shape changes depending on where the Moon is, and, worse, the oceans slop around in complicated ways depending on where the Moon is. And all these things alter $$g$$. Measuring the mass of the Moon this way is hopeless, sadly.

But, but. Could you even detect the Moon was there this way? One approach would be to have some device which measured $$g$$, over long periods of time, and then see if you could see periodic changes in $$g$$ which corresponded to the Moon.

Well, such a device is called a clock: a pendulum clock is sensitive to $$g$$, and if you made a really good pendulum clock would it 'hear' the Moon? The answer is yes, it would. And in 1986 someone called Boucheron did this experiment: they recorded the timekeeping of a very good pendulum clock (Shortt number 41) for almost a year and it was then analysed, first by Boucheron and then later by Philip Woodward. And if you look at the frequency spectrum of the clock you can see, very clearly, a number of spikes which correspond to the Sun and the Moon.

Unfortunately there are two problems with this wonderful experiment:

• you need to be able to measure the timekeeping of the clock with something more accurate than the clock – this was possible in 1986 using an atomic clock, but not possible when these clocks were the best clocks that existed;
• the clock can 'hear' the Moon, but it doesn't distinguish between the change in $$g$$ due to the direct gravitational effect of the Moon and that due to tides and deformation of the Earth.

This experiment is reported in My own right time by Philip Woodward, which has references to the original papers (which I have not read).

So, in summary: measuring the mass of the Moon by measuring $$g$$ was not plausible when it was one of the ways you might choose to measure the mass of the Moon, because the accuracy needed was too great.

• The Earth is spherical, just not perfectly and, as you write, it gets stretched by tidal forces a tiny little bit. Jun 21, 2020 at 13:43
• @LoveForChrist: the 'not perfectly' bit matters, as does the deformation (which is significant fractions of a metre)
– user21103
Jun 21, 2020 at 13:58
• When we talk of the shape of a planet we usually mean the one according to (what is considered the) sea level, or altitude 0. So by excluding mountains and such, the Earth is an oblate spheroid (because it is stretched by rotational forces) and additionally stretched by what you describe. Jun 21, 2020 at 14:49