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प्रश्न
The mass of planet ‘X” is four times that of the earth and its radius is double the radius of the earth. The escape velocity of a body from the earth is 11.2 × 103 m/s. Find the escape velocity of a body from the planet 'X’.
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उत्तर
Given: Escape velocity on earth’s surface (vesc) = 11.2 × 103 m/s,
Ratio of Planet (X) and earth’s mass (MX/Me) = 4,
Ratio of Planet (X) and earth’s radius (RX/Re) = 2
To find: Escape velocity (ve)X
Formulae:
- `"V"_"esc" = sqrt((2"GM"_"e")/"R"_"e")`
- `("V"_"esc")_"X" = sqrt((2"GM"_"X")/"R"_"X")`
Calculation: From formula (i) and (ii),
`("V"_"esc")_"X"/"V"_"esc" = sqrt(("M"_"X" xx "R"_"e")/("M"_"e" xx "R"_"X"))`
`= sqrt(4 xx 1/2)`
= 1.414
∴ (vsec)x = vesc × 1.414
= 11.2 × 103 × 1.414
= 15.84 × 103 m/s
The escape velocity of a body from the planet ‘X’ is 15.84 × 103 m/s.
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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.
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The stone held in the hand is stable because on it __.
- two unbalanced forces are exerted.
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