the radius of earth is 6370 . an object 30kg is taken to a height 230km above the surface of earth . what is the mass of the object . what is the acceleration . what is the weight ???
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Answer:
The radius of the Earth is about 6370 Km. An object of mass 30 Kg is taken to a height of 230 Km above the surface of Earth. What is the acceleration to gravity at this height? What is the weight of the body at this height?
From Newton’s Universal Law of Gravitation, g = G * M / R^2 where G is the universal constant, M is the mass of the planet and R^2 is the square of the distance from the planet’s center. If the mass go to 230 km from the surface of the planet Earth then its distance from the center of the Earth is (6370 Km + 230 km) = 6600 km. R^2 then becomes (R + 230)^2. The G will not be affected but since R^2 was affected the value of g is also affected.
What is the ratio of the former g1 to the new g2 ?. The ratio of g1 to g2 is equal to (R / R + 230)^2.
Solving for the new g2
g2 = g1 * (R / R + 230)^2
g2 = 9.8 m/s^2 * (6370 / 6370 + 230)^2
g2 = 9.8 m/s^2 * (6370 / 6600)^2
g2 = 9.8 m/s^2 * 0.931517
g2 = 9.1289 m/s^2 is the acceleration due to gravity at 230 km above the surface
The weight of the 30 kg body is Weight = mass * g2 or weight = 30 kg * 9.1289 m/s^2. Its weight is 273.9 Newtons.