Consider a launch vehicle of mass 10 tons being launched vertically. The vehicle has 8 tons of propellant, which burns completely at a constant rate over 50 s. If the engine specific impulse is 250 s, and the acceleration due to gravity at sea level is \(g_0\), the acceleration experienced by the vehicle at lift-off is ________.
Show Hint
- \(g_0\)
- \(2g_0\)
- \(3g_0\)
- \(4g_0\)
The Correct Option is C
Solution and Explanation
The vehicle burns 8 tons of propellant at a constant rate over 50 s. Using 1 ton = 1000 kg, that is 8000 kg burned in 50 s, so
\[ \dot{m} = \frac{8000}{50} = 160 \text{ kg/s} \]
Step 2: Find the thrust from the specific impulse.
Specific impulse links thrust, mass flow rate, and \(g_0\) as
\[ I_{sp} = \frac{F}{\dot{m}g_0} \implies F = \dot{m}\,I_{sp}\,g_0 \]
\[ F = 160 \times 250 \times g_0 = 40000\,g_0 \text{ N} \]
Step 3: Apply Newton's second law at the instant of lift-off.
At lift-off, none of the propellant has burned yet, so the vehicle still has its full initial mass, \(m_0=10\) tons \(=10000\) kg. Two forces act on it: the upward thrust \(F\) and the downward weight \(m_0g_0\). The net upward force gives the net acceleration:
\[ a = \frac{F - m_0g_0}{m_0} = \frac{F}{m_0} - g_0 \]
Step 4: Substitute the numbers.
\[ \frac{F}{m_0} = \frac{40000\,g_0}{10000} = 4g_0 \]
\[ a = 4g_0 - g_0 = 3g_0 \]
Step 5: Check the other options.
Option (D), \(4g_0\), is the thrust-to-mass ratio \(F/m_0\) on its own, before subtracting the vehicle's own weight; it is not the net acceleration the vehicle actually feels, since gravity is still pulling it down throughout. Options (A) and (B) do not come from a correct force balance at all: skipping the weight term or using the wrong mass gives numbers that are too small.
Final Answer:
The vehicle accelerates upward at \(3g_0\) at the moment of lift-off. \[ \boxed{a = 3g_0} \]
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