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8.17 A rocket is fired vertically with a speed of 5 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; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^–11 N m2 kg^–².

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Vishal Baghel | Contributor-Level 10

4 months ago

Velocity of the rocket, v = 5 km/s = 5 $* 10^{3}$ m/s

Mass of the Earth, $M_{e}$ = 6.0 * 10²⁴kg

Radius of the Earth, $R_{e}$ = 6.4 * 10⁶ m

Height reached by rocket mass, m = h

At the Earth's surface

Total energy of the rocket = Kinetic energy + Potential energy = $\frac{1}{2}$ m $v^{2}$ + ( $\frac{- G {m M}_{e}}{R_{e}}$ )

At highest point h, v = 0, and potential energy = $\frac{- G {m M}_{e}}{R_{e} + h}$

Total energy of the rocket at height h = $\frac{- G {m M}_{e}}{R_{e} + h}$

From the law of conservation of energy, we have,

Total energy of the rocket at Earth surface = Total energy at height h

$\frac{1}{2}$ m $v^{2}$ + ( $\frac{- G {m M}_{e}}{R_{e}}$ ) = $\frac{- G {m M}_{e}}{R_{e} + h}$ or $\frac{1}{2} v^{2}$ = $G M_{e} (\frac{1}{R_{e}} - \frac{1}{R_{e} + h})$

$\frac{1}{2} v^{2}$ = $G M_{e} (\frac{1}{R_{e}} - \frac{1}{R_{e} + h})$ = $\frac{G M_{e} h}{R_{e} (R_{e} + h)}$ = $\frac{G M_{e}}{R_{e}^{2}} * \frac{R_{e} h}{(R_{e} + h)}$ = $\frac{{g R}_{e} h}{(R_{e} + h)}$

...more

Velocity of the rocket, v = 5 km/s = 5 $* 10^{3}$ m/s

Mass of the Earth, $M_{e}$ = 6.0 * 10²⁴kg

Radius of the Earth, $R_{e}$ = 6.4 * 10⁶ m

Height reached by rocket mass, m = h

At the Earth's surface

Total energy of the rocket = Kinetic energy + Potential energy = $\frac{1}{2}$ m $v^{2}$ + ( $\frac{- G {m M}_{e}}{R_{e}}$ )

At highest point h, v = 0, and potential energy = $\frac{- G {m M}_{e}}{R_{e} + h}$

Total energy of the rocket at height h = $\frac{- G {m M}_{e}}{R_{e} + h}$

From the law of conservation of energy, we have,

Total energy of the rocket at Earth surface = Total energy at height h

$\frac{1}{2}$ m $v^{2}$ + ( $\frac{- G {m M}_{e}}{R_{e}}$ ) = $\frac{- G {m M}_{e}}{R_{e} + h}$ or $\frac{1}{2} v^{2}$ = $G M_{e} (\frac{1}{R_{e}} - \frac{1}{R_{e} + h})$

$v^{2} (R_{e} + h)$ = 2 ${g R}_{e} h$ $v^{2} R_{e} + v^{2} h$ = 2 ${g R}_{e} h$ h(2 ${g R}_{e} - v^{2})$ = $v^{2} R_{e}$

h = $\frac{v^{2} R_{e}}{(2 {g R}_{e} - v^{2})}$ = 1.59 x $10^{6}$ m

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8.17 A rocket is fired vertically with a speed of 5 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; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^–11 N m2 kg^–².

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8.17 A rocket is fired vertically with a speed of 5 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; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.

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8.17 A rocket is fired vertically with a speed of 5 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; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^–11 N m2 kg^–².