Earth's average radius is \(6,370 \mathrm{km}\) and its mass is \(5.97 \times 10^{24} \mathrm{kg}\) Show that the acceleration of gravity at the surface of Earth is \(9.81 \mathrm{m} / \mathrm{s}^{2}\).

Earth's mass is a crucial component when calculating the gravitational acceleration at the surface of the Earth. The mass of the Earth is a staggering value of \(5.97 \times 10^{24} \mathrm{kg}\). This massive amount of matter contributes to the gravitational pull we experience on the surface.

Understanding this value helps explain why gravity is strong enough to hold the atmosphere in place and keep us grounded. It is this immense mass that generates the gravitational force according to Newton's law of universal gravitation.

The Earth's radius is another essential factor in determining the gravitational acceleration. The average radius of the Earth is \(6,370 \mathrm{km}\), which converts to \(6.37 \times 10^6 \mathrm{m}\).

This radius is vital because gravitational force decreases with the square of the distance from the object's center of mass. By using the Earth's radius in our calculations, we can determine the gravitational acceleration on the surface.

The greater the radius, the smaller the gravitational pull at the surface, assuming the mass stays constant. This is inversely proportional to the square of the radius (\( R^2 \)) in the gravitational acceleration formula.

The gravitational constant, denoted as \(G\), is a fundamental value in physics that appears in the equation for gravitational force. The value of \(G\) is \(6.674 \times 10^{-11} \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}\). This constant allows us to relate the gravitational force to the masses involved and the distance between them.

Incorporating \(G\) in calculations with Earth's mass and radius lets us derive the gravitational acceleration at the Earth's surface. The formula used is \[ g = \frac{G M}{R^2} \].

By substituting Earth's mass and radius into this equation, and applying the gravitational constant \(G\), we can accurately compute the acceleration due to gravity as \(9.81 \mathrm{m} / \mathrm{s}^2 \). This acceleration is constant and plays a fundamental role in everyday life, from falling objects to the mechanics of orbital satellites.