All the time, the acceleration of gravity on the surface of the celestial body is g and the radius of the celestial body is r

All the time, the acceleration of gravity on the surface of the celestial body is g and the radius of the celestial body is r

Simple. M = GR ^ 2 / G is obtained from g = GM / R ^ 2
Where G is the gravitational constant. The sphere product v = 4 / 3 * π R ^ 3,
Density ρ = m / v = 3G / (4 π RG)

If the radius of a celestial body is R and the density is p, how many meters per square second is the acceleration of gravity g on its surface?

According to the law of universal gravitation: F = GMM / R ^ 2 density p = m / v = m / (4 / 3 * R ^ 2) simultaneous: F = 3 / 4 * GM g = f / M = 3G / 4 g is the constant of universal gravitation

Determination of gravitational acceleration on celestial body surface by the method of near earth satellite orbit

The calculation method of celestial gravity acceleration is as follows
Because f = GMM / R ^ 2, f = g = mg
So g = GM / R ^ 2
G: Gravitational constant = 6.67259 * 10 ^ - 11nm ^ 2 / kg ^ 2 (m ^ 3 / K · GS ^ 2)
M: Mass of central celestial body / kg
r: Distance between celestial body center and object center / M
The unit of G is m / S ^ 2 or N / kg