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प्रश्न
Calculate the escape velocity on the surface of the moon given the mass and radius of the moon to be 7.34 × 1022 kg and 1.74 × 106 m respectively.
(Given: G = 6.67 × 10-11 Nm2/kg2)
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उत्तर
Given:
G = 6.67 × 10-11 Nm2/kg2
Mass of the moon = M = 7.34 × 1022 kg
The radius of the moon = R = 1.74 × 106.
Escape velocity = Vesc = `sqrt((2"GM")/"R")`
= `sqrt((2xx6.67xx10^-11xx7.34xx10^22)/(1.74xx10^6))`
= 2.37 km/s
The escape velocity on the moon is 2.37 km/s.
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संबंधित प्रश्न
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Read the given passage carefully and answer the questions.
We know that the force of gravity due to the earth acts on each and every object. When we were holding the stone in our hand, the stone was experiencing this force, but it was balanced by a force that we were applying on it in the opposite direction. As a result, the stone remained at rest. Once we release the stone from our hands, the only force that acts on it is the gravitational force of the earth and the stone falls down under its influence. Whenever an object moves under the influence of the force of gravity alone, it is said to be falling freely. Thus the released stone is in a free fall. In free fall, the initial velocity of the object is zero and goes on increasing due to the acceleration due to the gravity of the earth. During free fall, the frictional force due to air opposes the motion of the object and a buoyant force also acts on the object. Thus, true free fall is possible only in a vacuum.
1) Complete the following statement by choosing the right option.
The stone held in the hand is stable because on it __.
- two unbalanced forces are exerted.
- only the gravitational force of the earth is exerted.
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