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Friday, 1 July 2011

Earth's gravity

Earth's gravity

Every planetary body (including the Earth) is surrounded by its own gravitational field, which exerts an attractive force on all objects. Assuming a spherically symmetrical planet (a reasonable approximation), the strength of this field at any given point is proportional to the planetary body's mass and inversely proportional to the square of the distance from the center of the body.
The strength of the gravitational field is numerically equal to the acceleration of objects under its influence, and its value at the Earth's surface, denoted g, is approximately expressed below as the standard average.
g = 9.81 m/s2 = 32.2 ft/s2
This means that, ignoring air resistance, an object falling freely near the Earth's surface increases its velocity by 9.81 m/s (32.2 ft/s or 22 mph) for each second of its descent. Thus, an object starting from rest will attain a velocity of 9.81 m/s (32.2 ft/s) after one second, 19.6 m/s (64.4 ft/s) after two seconds, and so on, adding 9.81 m/s (32.2 ft/s) to each resulting velocity. Also, again ignoring air resistance, any and all objects, when dropped from the same height, will hit the ground at the same time.
If an object with comparable mass to that of the Earth were to fall towards it, then the corresponding acceleration of the Earth really would be observable.
According to Newton's 3rd Law, the Earth itself experiences a force equal in magnitude and opposite in direction to that which it exerts on a falling object. This means that the Earth also accelerates towards the object until they collide. Because the mass of the Earth is huge, however, the acceleration imparted to the Earth by this opposite force is negligible in comparison to the object's. If the object doesn't bounce after it has collided with the Earth, each of them then exerts a repulsive contact force on the other which effectively balances the attractive force of gravity and prevents further acceleration.

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