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The Hubble telescope’s orbit is 5.6 x 105 meters above Earth’s surface. The telescope has a mass of 1.1 x 104 kilograms. Earth exerts a gravitational force of 9.1 x 104 newtons on the telescope. What is the magnitude of Earth’s gravitational field strength at this location?

Respuesta :

MissPhiladelphia
F=Gm1m2/r^2where:
F is the force between the masses;G is the gravitational constant ( N · (m/kg)2);m1 is the first mass;m2 is the second mass;r is the distance between the centers of the masses.
m1 = mass earth = 5.972 × 10^24 kg
9.1 x 10^4 N = G(5.972 × 10^24 kg)(1.1 x 10^4 kg)/(5.6 x 10^5 m)^2 solve for GG = 4.34x10^-13 N · (m/kg)2

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