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Maharashtra State BoardSSC (English Medium) 10th Standard

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

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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
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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

`"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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Chapter 1: Gravitation - Solve the following Questions [Page 25]

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SCERT Maharashtra Science and Technology Part 1 [English] Standard 10 Maharashtra State Board
Chapter 1 Gravitation
Solve the following Questions | Q 7. | Page 25

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