why is acceleration due to gravity constant
Equations for free fall can be used for calculating acceleration due to gravity. The equation,
h = gt 2 can be solved for g, as, g Experiment Conducted to find Acceleration due to gravity: An object of considerable weight (because the air resistance can be neglected), be dropped from fairly tall building whose height is accurately measured. The observer at the top of the building switches on a digital timer at the same instant he drops the object. The observer at ground level switches off the same timer when the object hits the ground and records the time of fall. Plugging the values of h and t, the acceleration due to gravity can be calculated.
The experiment may be conducted for a few times and the mean of the readings may be taken. Bodies allowed to fall freely were found to fall at the same rate irrespective of their masses (air resistance being negligible). The velocity of a freely falling body increased at a steady rate, i. e. , the body had acceleration. We know, From Newtons Second Law, F = mg. Newton's law of gravity, states that every point mass in the universe is attracts to every other point mass in the universe with a force which is directly proportional to the product of their masses and inversely proportional to square of the distance between them. It can be written as, F. F is the force, m is the mass of the body, g is the acceleration due to gravity, M is the mass of the Earth, R G By knowing these constants, we can calculate "g" theoretically.
From equations (1) and (2), we can also conclude that 'g' varies with (c) Latitude. 1) The radius of the moon is 1. 74 x 10 m. The mass of the moon is 7. 35 x 10 kg. Find the acceleration due to gravity on the surface of the moon. On the surface of the moon, the distance to the center of mass is the same as the radius: r = 1. 74 x 10 m = 1 740 000 m. The acceleration due to gravity on the surface of the moon can be found using the formula: g = 1. 620 m/s The acceleration due to gravity on the surface of the moon is 1. 620 m/s. 2) The radius of the Earth is 6. 38 x 10 m.
The mass of the Earth is 5. 98x 10 kg. If a satellite is orbiting the Earth 250 km above the surface, what acceleration due to gravity does it experience? The acceleration due to gravity depends on the distance from the center of mass of the large body to the satellite. This distance is the sum of the radius of the Earth and the distance from the satellite to the surface: r =(6. 38 x 10 r r g = 9. 078 m/s The acceleration due to gravity at the height of the satellite, 250 km above the surface of the Earth, is 9. 078 m/s.
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