And it could be done, in principle, using methods available to the ancient Greeks, so long as you are allowed to have knowledge of the gravitational constant, [math]G=6.674\times 10^{-11}~{\rm m}^3/{\rm kg}{\rm s}^2. CALCULATING THE MASS OF THE EARTH 1. To calculate the mass of the Earth, all one needs to do is divide by $G$. This lack of knowledge of $G$ inherently plagues any precise measurement of the mass of the Earth.
It is given as the force (F), which is equal to the Gravitational constant multiplied by the mass of the planet and the mass of the object, divided by the square of the radius of the planet. But it certainly is not massless. By measuring the attractive force between spheres due to gravity, he was able to use that to determine the density of the earth and therefore the earth’s mass. Looking at pendulum deflection, you can calculate the ratio of the mass of the Earth to the mass of Schiehallion. Consider a body of mass m on the surface of the earth. You ask about the mass of the Earth and not about its weight.
If the product is $G M_E$ is known to a high degree of accuracy (and it is), dividing by $G$ will lose a lot of accuracy because the gravitational constant $G$ is only known to four decimal places of accuracy. The sidereal period of the moon, which is 27.3 days, will give you a calculation of Earth's mass that's more accurate than the calendar period of the moon. Where M e is the mass of the Earth, in kilograms, v is the average velocity of the moon, r is the average distance between the moon and the Earth. According to the law of gravitation, the gravitational force F of the earth acting on a body is given by: =6 × 10 24 kg Using a modern Digital Terrain Model and geological models, Maskelyn's pendulum measurements give a result that agrees with the current accepted value of G (or M - … The study of the size and shape of the earth is known as geodesy and has been practiced for centuries. Technically Earth's shape is called the "geoid", an ellipsoidal shape. The centripetal force is the Earth's mass times the square of its speed divided by its distance from the sun. There's a catch, though. This procedure was actually used to calculate the mass of the earth and it is used to calculate, well, the mass of any orbited body. What is that magic equation? 2.
This is an old one, but it’s popped up as somebody followed it recently, and so I guess it’s come around again in exam revision. There are a couple of correct answers already, but I’ll try to provide something a little clearer. In fact, we did not know the mass of the planet Mercury until we put a satellite around it. F = GmM/r 2 = ma, where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, r is the radius of the Earth, and m is the mass of another object (near the surface of the Earth). The mass of the Earth may be determined using Newton's law of gravitation. Let the mass of the earth be M and radius of the earth be R, The distance of the body from the center of the earth will also be equal to the radius R of the earth. I suppose that is because you already know that as the Earth orbits the Sun it is weightless. Sure it is. Well, there are actually several. Knowing this value of g for Earth's surface, along with the constant G and the 6,731-kilometer distance to Earth's center, you can then calculate Earth's mass to be 6 x 10 24 kilograms. I suppose that is because you already know that as the Earth orbits the Sun it is weightless. Contrary to common misconceptions, many historical mathematicians and scientists were aware that Earth was spherical. and G is the universal gravitation constant.
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