A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth? (G = 6.67 × 10^-11N ∙ m²/kg², Mearth = 5.97 × 10²⁴ kg, the polar radius of the earth is 6357 km) A) 4.5 km/s B) 2.2 km/s C) 2.7 km/s D) 3.2 km/s E) 4.8 km/s

Question

Quizplus · Accepted Answer

The correct answer is A because the object is released from rest at an altitude of 1200 km above the north pole of the earth, and we ignore atmospheric friction. So the object will fall freely under the gravity of the earth. The speed of the object when it strikes the surface of the earth can be calculated by using the conservation of energy. The initial energy of the object is all potential energy, and the final energy of the object is all kinetic energy. So we have:$$mgh = \frac{1}{2}mv^2$$where m is the mass of the object, g is the acceleration due to gravity, h is the altitude of the object, and v is the speed of the object.Solving for v, we get:$$v = \sqrt{2gh}$$Substituting the given values, we get:$$v = \sqrt{2 	imes 6.67 	imes 10^{-11} 	imes 5.97 	imes 10^{24} 	imes 1200000} = 4.5 	ext{ km/s}$$

A 910-Kg Object Is Released from Rest at an Altitude