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
m = 4.0 * pi*pi * radius*radius*radius / ( period*period * G)
return m
def gravity(mass, radius): # calculate planet surface gravity from mass and radius
g = G * mass / (radius*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.
See also: