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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. - Science and Technology 1

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Question

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’. 

Numerical

Solution

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: 

  1. `"V"_"esc" = sqrt((2"GM"_"e")/"R"_"e")`
  2. `("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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Chapter 1: Gravitation - Solve the following Questions

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SCERT Maharashtra Science and Technology 1 [English] 10 Standard SSC
Chapter 1 Gravitation
Solve the following Questions | Q 7

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