हिंदी

The escape velocity of a body on a planet 'A' is 12 kms–1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is ______. -

Advertisements
Advertisements

प्रश्न

The escape velocity of a body on a planet 'A' is 12 kms–1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is ______.

विकल्प

  • 12 kms-1

  • 24 kms-1

  • 36 kms-1

  • 6 kms-1

MCQ
योग

उत्तर

The escape velocity of a body on a planet 'A' is 12 kms–1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is 12 kms-1.

Explanation:

Escape velocity of a body on a planet is given by:

Ve = `sqrt(2"gR")` or V= `sqrt(2"GM")/"R"`

where g ⇒ acceleration due to gravity at the surface

m ⇒ mass of planet and R = Radius of planet

Assuming plant to be solid sphere uniform, density ρ

Mass = Density × Volume

M = ρ × `4/3` πR3

The expression for escape velocity can be written as

V= `sqrt (2  rho xx -  pi"R")/"R"` 

= `sqrt(4/3"G"rho"R")`

V= for planet A = 12 km/s.

Let it’s Radius RA and density ρ

For planet B ⇒ `("V"_e)_"B" ⇒ "R"_"B"`

= `("R"_A")/2`, ρB = 4 ρA

`(("V"_"e")_"A")/(("V"_"e")_"B") = sqrt((8pi  "G"rho_"A""R"_"A"^2)/3)/((8pi  "G"rho_"B""R"_"B"^2)/3)`

= `sqrt(rho_"A"/rho_"B" ("R"_"A"/("R"_"B"))^2`

= `sqrt(1/4xx4)`

= 1

`("V"_"e")_"A"` = `("V"_"e")_"B"` = 12 km/s

shaalaa.com
Introduction to Gravitation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×