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Exercises · 7.17

Q.A rocket is fired vertically with a speed of 5 km s−15\text{ km s}^{-1} from the earth's surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth =6.0×1024 kg= 6.0 \times 10^{24}\text{ kg}; mean radius of the earth =6.4×106 m= 6.4 \times 10^{6}\text{ m}; G=6.67×10−11 N m2 kg−2G = 6.67 \times 10^{-11}\text{ N m}^{2}\text{ kg}^{-2}.

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By energy conservation the rocket's launch kinetic energy converts fully into gravitational potential energy. It reaches a distance rmax⁡=8.0×106 mr_{\max} = 8.0 \times 10^{6}\ \text{m} from the earth's centre, i.e. a height h=1.6×106 mh = 1.6 \times 10^{6}\ \text{m} (1600 km) above the earth's surface.

Concept

Gravity is conservative, so total mechanical energy is conserved between launch (speed vv, at the surface r=Rr=R) and the highest point (speed 00, at r=rmax⁡r = r_{\max}). The gravitational potential energy of a mass mm at distance rr from the earth's centre is U=−GMmrU = -\dfrac{GMm}{r}.

Energy conservation

12mv2−GMmR=−GMmrmax⁡\frac{1}{2}mv^{2} - \frac{GMm}{R} = -\frac{GMm}{r_{\max}}

The rocket's mass mm cancels, so the result is independent of the rocket's mass:

GMrmax⁡=GMR−12v2⇒rmax⁡=GMGMR−12v2\frac{GM}{r_{\max}} = \frac{GM}{R} - \frac{1}{2}v^{2} \qquad\Rightarrow\qquad r_{\max} = \frac{GM}{\dfrac{GM}{R} - \dfrac{1}{2}v^{2}}

Substitute the values

v=5 km s−1=5000 m s−1v = 5\ \text{km s}^{-1} = 5000\ \text{m s}^{-1}, M=6.0×1024 kgM = 6.0\times10^{24}\ \text{kg}, R=6.4×106 mR = 6.4\times10^{6}\ \text{m}, G=6.67×10−11 N m2kg−2G = 6.67\times10^{-11}\ \text{N m}^{2}\text{kg}^{-2}.

GM=(6.67×10−11)(6.0×1024)=4.002×1014 m3 s−2GM = (6.67\times10^{-11})(6.0\times10^{24}) = 4.002\times10^{14}\ \text{m}^{3}\,\text{s}^{-2}

GMR=4.002×10146.4×106=6.253×107 J kg−1\frac{GM}{R} = \frac{4.002\times10^{14}}{6.4\times10^{6}} = 6.253\times10^{7}\ \text{J kg}^{-1} …

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