Question

A projectile is projected with a speed $K{V_e}$ where ${V_e}$ is escape velocity and K is a constant less than one. The maximum height reached by it from the center of earth will be.
1)$\dfrac{R}{{{K^2} + 1}}$;
2) $\dfrac{R}{{{K^2} - 1}}$
3) $\dfrac{{R{K^2}}}{{1 - {K^2}}}$
4) $\dfrac{{{K^2} - 1}}{R}$

Answer 1

Hint: Here there is no net external force so one can apply the conservation of energy in the question. Here the total initial energy would be equal to the total final energy. Apply this method and find the height.

Complete answer:
Apply conservation of energy and solve:
${\text{Total Energy = }}KE + PE$;
Now the work done to throw a particle:
$W = F.dx$; ….(dx = r)
$ \Rightarrow $$W = F.r$; …(Here F = Force due to gravity)
$ \Rightarrow $$W = - \dfrac{{(GMm)}}{{{r^2}}}.r$;
$ \Rightarrow $$W = - \dfrac{{GMm}}{r}$;
Here the work done due to gravity is equal to the potential energy.
${U_g} = - \dfrac{{GMm}}{r}$; …(Initial P.E)
For initial kinetic energy;
$K.E = \dfrac{1}{2}m{K^2}{v_e}^2$; …(Here k is constant and ${v_e}$is escape velocity)
So the initial total energy becomes:
${\text{TE = }}\dfrac{1}{2}m{k^2}v_e^2 - \dfrac{{GMm}}{r}$;
Similarly the final total energy will become:
${\text{TE = K}}{\text{.E}} - \dfrac{{GMm}}{{r + h}}$; …(Here: h is the height above the earth’s surface)
Now according to the conservation of energy the initial energy will be equal to the final energy:
$\dfrac{1}{2}m{k^2}v_e^2 - \dfrac{{GMm}}{r}{\text{ = }}\left( {{\text{K}}{\text{.E - }}\dfrac{{GMm}}{{r + h}}} \right)$;
In the above equation for the final energy K.E will be zero :
$\dfrac{1}{2}m{k^2}v_e^2 - \dfrac{{GMm}}{r} = - \dfrac{{GMm}}{{r + h}}$;
Put the formula for escape velocity in the above equation:
${V_e} = \sqrt {2gr} $;
$ \Rightarrow $$ - \dfrac{{GMm}}{r} + \dfrac{1}{2}m{k^2} \times 2gr = - \dfrac{{GMm}}{{r + h}}$;
Simplify the above equation:
\[m{k^2}gr = \dfrac{{GMm}}{r} - \dfrac{{GMm}}{{r + h}}\];
$ \Rightarrow $\[m{k^2}gr = GMm\left( {\dfrac{1}{r} - \dfrac{1}{{r + h}}} \right)\];
Cancel out the common factors:
\[{k^2}gr = GM\left( {\dfrac{1}{r} - \dfrac{1}{{r + h}}} \right)\];
Take the LCM on the RHS;
\[{k^2}gr = GM\left( {\dfrac{1}{r} - \dfrac{1}{{r + h}}} \right)\];
$ \Rightarrow $\[{k^2}gr = GM\left( {\dfrac{{r + h - r}}{{r\left( {r + h} \right)}}} \right)\];
Simplify the equation:
\[{k^2}gr = GM\left( {\dfrac{h}{{\left( {{r^2} + hr} \right)}}} \right)\];
Take ${r^2}$out as common on the denominator on LHS.
\[{k^2}gr = GM\left( {\dfrac{h}{{{r^2}\left( {1 + \dfrac{h}{r}} \right)}}} \right)\];
$ \Rightarrow $\[{k^2}gr = \dfrac{{GM}}{{{r^2}}}\left( {\dfrac{h}{{\left( {1 + \dfrac{h}{r}} \right)}}} \right)\];
Here$g = \dfrac{{GM}}{{{r^2}}}$;
$ \Rightarrow $\[{k^2}r = \left( {\dfrac{h}{{\left( {1 + \dfrac{h}{r}} \right)}}} \right)\];
$ \Rightarrow $\[{k^2}r = \dfrac{h}{{\left( {1 + \dfrac{h}{r}} \right)}}\];
Take the value on RHS in the denominator and put it in LHS:
\[{k^2}r \times \left( {1 + \dfrac{h}{r}} \right) = h\];
$ \Rightarrow $\[{k^2}r + {k^2}h = h\];
Take \[{k^2}h\]on the RHS:
\[{k^2}r = h - {k^2}h\];
$ \Rightarrow $\[{k^2}r = h\left( {1 - {k^2}} \right)\];
Solve for h:
\[\dfrac{{r{k^2}}}{{\left( {1 - {k^2}} \right)}} = h\]; …(r = R)

Option”3” is correct. The maximum height reached by it from the center of earth will be \[\dfrac{{r{k^2}}}{{\left( {1 - {k^2}} \right)}} = h\].

Note:
The process is complex so go step by step make sure there are no mistakes while simplifying the equation. Here in the final energy the kinetic energy would be equal to zero as at maximum height there is no kinetic energy.