Question
Download Solution PDFA particle of mass ‘m’ is projected with a velocity v=kVe(k < 1) from the surface of the earth. (Ve=escape velocity) The maximum height above the surface reached by the particle is :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The minimum velocity required by a body to overcome the gravitational force of earth and escape from earth so that it never falls back is called as escape velocity.
Escape velocity formula is given
\({V} = \sqrt {\frac{{2GM}}{R}} \)
Where, V is the escape velocity
G is the gravitational constant is 6.67 × 10-11 m3 kg-1 s-2
M is the mass of the planet
R is the radius from the center of gravity
Calculation:
Given v = kVe
where, k < 1
Thus, v < Ve
From conservation of mechanical energy,
\(\frac{1}{2}m{v^2} - \frac{{GmM}}{R} = - \frac{{GmM}}{{(R + h)}}\)
\( \Rightarrow \frac{{{V^2}}}{2} = \frac{{GM}}{R} - \frac{{(GM)}}{{(R + h)}} = \frac{h}{{R(R + h)}}GM\)
\( \Rightarrow \frac{1}{2}{k^2}V_e^2 = \frac{{GMh}}{{R(R + h)}}\)
We know, \({V_e} = \sqrt {\frac{{2GM}}{R}} \)
\( \Rightarrow \frac{1}{2}{k^2}\left( {\frac{{2GM}}{R}} \right) = \frac{{GMh}}{{R(R + h)}}\)
\({k^2} = \frac{h}{{(R + h)}}\)
Rk2 + hk2 = h
Rk2 = h(1 – k2)
\(\therefore h = \frac{{R{k^2}}}{{(1 - {k^2})}}\)
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