Gone Forever

Classical Mechanics Level 3

I toss a stone at a velocity of 14000 m/s 14000 \text{ m/s} , and an angle chosen uniformly at random from the interval [ 0 ∘ , 9 0 ∘ ] [0^\circ, 90^\circ] , above the horizontal. What is the probability that the stone will never return to Earth?

Details and Assumptions :

  • g = 9.8 m/s 2 g=9.8 \text { m/s}^2 , radius of Earth 6371 km.

  • Ignore other massive bodies, such as the sun.


The answer is 1.000.

Alex Li by Alex Li , 22, USA

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1 solution

Alex Li
Apr 24, 2015

The escape velocity is given by 2 g R = 2 × 9.8 × 6371000 = 11175 \sqrt{2gR}=\sqrt{2\times9.8\times6371000}=11175 m / s m/s . Because the escape velocity doesn't depend on the initial angle of projection, and the stone's velocity is greater than the escape velocity, the stone is guaranteed to escape. Thus, the probability is 1 \boxed{1} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Shouldnt we be considering just the vertical component of the velocity of projection? I don't think the horizontal component really plays a part in helping the stone escape. The formula for escape velocity also is derived considering vertical projection. Or am I going wrong somewhere?

Utkarsh Singh - 6 years, 1 month ago

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