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  • Confused! kindly explain, - Two stars of masses each, and at distance rotate in a plane about their common centre of mass O. A - Gravitation - JEE Main

Two stars of masses 3\times 10^{31}kg each, and at distance 2\times 10^{11}mrotate in a plane about their common centre of mass O. A meteorite passes though O moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is :

(Take Gravitational constant

G= 6.67 X 10-11 Nm2 kg -2)

 

  • Option 1)

    3.8\times 10^{4}\: m/s

  • Option 2)

    1.4\times 10^{5}\: m/s

  • Option 3)

    2.8\times 10^{5}\: m/s

  • Option 4)

    2.4\times 10^{4}\: m/s

Answers (1)

best_answer

 

Escape velocity -

Minimum velocity which is required to escape the body from earth's gravitational pull

- wherein

It is independent of the mass and direction of projection of body.

 

 

Escape velocity ( in terms of radius of planet) -

V_{c}=\sqrt{\frac{2GM}{R}}

V_{c}=\sqrt{2gR}

V_{c}\rightarrow Escape velocity

R\rightarrowRadius of earth

- wherein

  • depends on the reference body
  • greater the value of \frac{M}{R} or \left ( gR \right ) greater will be the escape velocity V_{e}=11.2Km/s  For earth

 

 

By energy conservation

-\frac{GMm}{r} - \frac{GMm}{r}+\frac{1}{2}mV^{2} = 0+0

M - mass of star

m - mass of meteroite

V = \sqrt{\frac{4Gm}{r}} = 2.8 \times 10^{5}m/s

 


Option 1)

3.8\times 10^{4}\: m/s

Option 2)

1.4\times 10^{5}\: m/s

Option 3)

2.8\times 10^{5}\: m/s

Option 4)

2.4\times 10^{4}\: m/s

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