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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
`"V"_"esc" = sqrt((2"GM"_"e")/"R"_"e")`
`("V"_"esc")_"X" = sqrt((2"GM"_"X")/"R"_"X")`
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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