A spaceship is fired from earth's surface with an initial speed of 15 km/sec, what will its speed be when it is very far from the earth? (neglect frictional resistance) $(v_{e} = 11 k m / s e c)$

4 km/sec

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10 km/sec

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8 km/sec

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7 km/sec

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Solution

The correct option is B 10 km/sec
Applying conservation of energy:
$\frac{1}{2} m V_{p r o j e c t i o n}^{2} - \frac{G M m}{R} = \frac{1}{2} m V_{\infty}^{2}$

but, we know that,
$\frac{G M m}{R} = \frac{1}{2} m V_{e s c a p e}^{2}$

$\Rightarrow V_{\infty} = \sqrt{(V_{p r o j e c t i o n})^{2} - (V_{e s c a p e})^{2}}$
$= \sqrt{15^{2} - 11^{2} =} \sqrt{225 - 121} = \sqrt{104} \approx \sqrt{100} = 10 k m / s e c$

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A spaceship is fired from earth's surface with an initial speed of 15 km/sec, what will its speed be when it is very far from the earth? (neglect frictional resistance) (ve=11 km/sec)

The correct option is B 10 km/secApplying conservation of energy: 12mV2projection−GMmR=12mV2∞ but, we know that, GMmR=12mV2escape ⇒V∞=√(Vprojection)2−(Vescape)2 =√152−112=√225−121=√104≈√100=10 km/sec

A spaceship is fired from earth's surface with an initial speed of 15 km/sec, what will its speed be when it is very far from the earth? (neglect frictional resistance) $(v_{e} = 11 k m / s e c)$