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Acceleration of gravity (g) universal gravitation

A body of mass m is attracted to Earth with a force that is worth its weight and P = mg. But under the law of universal gravitation, this force is also true: F = GmM/r2

, where M is the mass of Earth and r is the distance from the center of the Earth to the body of mass m. These forces being equal, is:

mM/r2 mg = G, ie g = GM/r2

Through this formula we can calculate the value of ga any distance from the center of the Earth, conocinedo M mass of it and the distance r.

This formula indicates that the farther from the center of the earth the acceleration of gravity is lower and, therefore, the bodies that they will fall uniformly accelerated motion, because as the body falls the value of g increases.

Invitation Death Anniversary

The law of universal gravitation was released by Isaac Newton in 1686 and the myth says it was inspired by an apple falling from a tree. His statement is:

"Every body in the universe attracts the other with a force that is proportional to the product of their masses and inversely proportional to the square of the distance that separates"

F = G m1 * M2 / r2, where G is a gravitational constant, put there by Cavendish (English, 1731-1810) using a torsion balance. G = 6.67 x10 (-11) (Mm2/Kg2) in MKS

This value of G means that two masses of 1 Kg 1 m are attracted by the power of 6.67 x10 (-11) N.

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Universal Gravitation Problems

1. Aceleracionde calculate gravity at the lunar surface if the dough is 1 / 81 the mass of Earth and its radius is 1738 km Mass of Earth = 6x10 (24) Kg
Answer: Done in class

2. At one point on Earth is a satellite of 500 kg on the Earth's gravitational field which acts with the force of 4000 Newton. What is the gravitational field strength and the acceleration of gravity at that distance?
A. Remember
gravitational field strength g = F / m, so 4000/500 = 8 N / kg .
The acceleration of gravity has the value mime 8 m/s2.

3. Calculate the gravitational field strength at 630 km from the earth's surface if the mass of Earth is 5.98 x 10 (24) kg and its radius 6370 km?
A. mM/r2
mg = G, the radius is 6370 km + 630 km = 7000 km, or 7 million m
is replaced in the equation and go: g = 8.14 m/s2 .

4. How far from the earth's surface rotates in a circular orbit sateliute if its mass is 1000 kg and the field acting on it with the force of 8000N?
A. Using F = GmM/r2, the radius is 6370 + d is squared. The 6370 km need to be transformed to meters. The result is d = 690 km approx.

Answer: 5. Calculate the gravitational field strength at a point located 3630 km from the earth's surface if the mass of Earth is 5.98 x 10 (24) kg and G = 6.67 x10-11
Answer:
The attractive force is equal weight, therefore we can equate P = G * mM/r2
if P = mg, mg = GmM/r2 say.
Masses are simplified and is g = GM/r2. replace the data, remember that the radius is 6370 km + 3630 = 10,000 km, that is 10 million meters from the center of the Earth. gives g = 4 N / kg or 4 m/s2 .

6. A 80-kg satellite rotates in a circular orbit. If the gravitational field acting on the satellite with a force of 16N. What is the gravitational field strength at that distance?
Answer:
gravitational field strength is expressed as g = F / m and its unit is N / Kg.
Then F = 16N m = 80kg, giving a g = 0.2 N / Kg .

Kepler

1. Revoluciuón period of Saturn is about 29.5 years. Calculate the distance from the sun. (Period of the earth = 1 and distance from the Sun = 1 AU) (1 AU = 1.5 x 10 (8) km)
Answer: Done in class.

2. The acceleration of gravity Mars relative to Earth is 0.37. Calculate the acceleration of gravity on Mars if Earth is m/s2 9.8 m/s2
Answer: aceleracionde
if Earth gravity is 9.8 m/s2 and that of Mars is 0.37 times that of Earth, then it says: g * g Mars = 0.37 Earth. The product gives a of 3.62 g on Mars m/s2.

3. The Earth's average density is 5.5 g/cm3 and that of Mars with relation to the Earth is 0.69. What is the density in g/cm3 Mars?
Answer:
take the density value of the Earth as 1 unit and that of Mars is 0.69 units. Then by rule 3 is obtained D = 3.79 g/cm3 Mars.

4. Calculate the acceleration of gravity at a point in the distance to the Moon is the Earth which is 60 Earth radii.
Answer:
g1 on the surface of the earth is GMt/r2 and g2 to the distance of the Moon is GMt / (60r) 2
by dividing both equations gives G1/G2 = (60r) 2/r2 = 3600 / 1
where g2 = 1 / 3600 * g1, g1 is taken as 9.8 m/s2 and you get g2 = 2.7 x10 (-3) m/s2.