Earth's mean radius and mass are 6,371 km and 5.97 x 1024 kg, respectively. Show that the acceleration of gravity at the surface of Earth is 9.80 m/s2.
You know the mass of earth, and a well known principle (accept it) that for symetrical distributions, you can consider that mass at the center of a sphere.
g=G*Mass/r^2
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To calculate the acceleration of gravity at the surface of Earth, we can use Newton's law of universal gravitation.
The formula for the acceleration of gravity is:
g = (G * M) / R^2
where:
g is the acceleration of gravity
G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2)
M is the mass of Earth
R is the radius of Earth
Given:
M = 5.97 x 10^24 kg
R = 6,371 km (or 6,371,000 m)
Now we can substitute the given values into the formula:
g = (6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.97 x 10^24 kg) / (6,371,000 m)^2
Calculating this expression, we get:
g = 9.80 m/s^2
Hence, we have shown that the acceleration of gravity at the surface of Earth is approximately 9.80 m/s^2.