Question
A particle is fired vertically upward from earth's surface and it goes up to a maximum height of 6400km. Find the initial speed of the particle.
✓
Answer
The particle attain maximum height = 6400km. On earth’s surface, its P.E. & K.E.$\text{E}_{\text{e}}=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{-\text{GMm}}{\text{R}}\Big)\ ...(1)$
In space, its P.E. & K.E.$\text{E}_\text{s}=\Big(-\frac{\text{GMm}}{\text{R}+\text{h}}\Big)+0$
$\text{E}_{\text{s}}=\Big(-\frac{\text{GMm}}{2\text{R}}\Big)\ ...(2)\ (\because\text{h}=\text{R})$
Equating (1) & (2)$-\frac{\text{Gmm}}{\text{R}}+\frac{1}{2}\text{mv}^2=-\frac{\text{Gmm}}{2\text{R}}$
or $\Big(-\frac{1}{2}\Big)\text{mv}^2=\text{GMm}\Big(-\frac{1}{2\text{R}}+\frac{1}{\text{R}}\Big)$ or $\text{v}^2=\frac{\text{GM}}{\text{R}}$$=\frac{6.67\times10^{-11}\times6\times10^{24}}{6400\times10^3}$
$=\frac{40.02\times10^{13}}{6.4\times10^6}$
$=6.2\times10^7=0.62\times10^8$
or $\text{v}=\sqrt{0.62\times10^8}=0.79\times10^4\text{m/s}=7.9\text{km/s.}$
In space, its P.E. & K.E.$\text{E}_\text{s}=\Big(-\frac{\text{GMm}}{\text{R}+\text{h}}\Big)+0$
$\text{E}_{\text{s}}=\Big(-\frac{\text{GMm}}{2\text{R}}\Big)\ ...(2)\ (\because\text{h}=\text{R})$
Equating (1) & (2)$-\frac{\text{Gmm}}{\text{R}}+\frac{1}{2}\text{mv}^2=-\frac{\text{Gmm}}{2\text{R}}$
or $\Big(-\frac{1}{2}\Big)\text{mv}^2=\text{GMm}\Big(-\frac{1}{2\text{R}}+\frac{1}{\text{R}}\Big)$ or $\text{v}^2=\frac{\text{GM}}{\text{R}}$$=\frac{6.67\times10^{-11}\times6\times10^{24}}{6400\times10^3}$
$=\frac{40.02\times10^{13}}{6.4\times10^6}$
$=6.2\times10^7=0.62\times10^8$
or $\text{v}=\sqrt{0.62\times10^8}=0.79\times10^4\text{m/s}=7.9\text{km/s.}$
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