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A 800 lb space probe is launched from Earth, headed for deep space.

a) List the correct equation and solve for mass given that the acceleration of gravity is 32.2 ft/s²

b) At a distance of 15,000 miles from the Earth's center, the space probe releases a payload that reduces the mass of the space probe by 1/4th (i.e., space probe is now 25% lighter in mass). By how much did the gravitational force go up or down after the payload has been released? Start with the equations from the text or class notes and solve.

Expert Solution

Part a

The rocket propulsion is based on the principle of conservation of momentum. As the fuel is expelled in opposite direction to rocket with relative exhaust velocity v_e with rocket itself moving at v m/s.

Consider the situation at say time t, the fuel and the probe both moving with velocity v, then

$P_{1} = (m_{p r o b e} + ∆ m_{f u e l}) v (1)$

Assuming the gas is expelled at a constant relative velocity v_e.

Now at a small interval dt later the gas is mass $∆ m_{f u e l}$ is expelled and the probe has mass m moving with velocity $v + ∆ v$ , the momentum of the system is now,

$P_{2} = m_{p r o b e} (v + ∆ v) + ∆ m_{f u e l} (v - v_{e}) (2)$

Before obtaining the probe equation one thing to note is that the

$∆ m_{f u e l} = - ∆ m_{p r o b e} (3)$

Using (3) in (2) gives,

$P_{2} = m (v + ∆ v) - ∆ m_{m_{p r o b e}} (v - v_{e}) (4)$

As the momentum of system is conserved (in absence of drag forces), this gives,

$P_{2} = P_{1} m_{p r o b e} (v + ∆ v) - ∆ m_{p r o b e} (v - v_{e}) = (m_{p r o b e} - ∆ m_{p r o b e}) v m_{p r o b e} ∆ v + ∆ m_{p r o b e} v_{e} = 0 dividing by time m_{p r o b e} d v + d m_{p r o b e} v_{e} = 0 rearranging gives, m_{p r o b e} d v = - v_{e} d m_{p r o b e} \int_{v}^{v + ∆ v} v = - v_{e} \int_{m_{i}}^{m_{f}} \frac{d m_{p r o b e}}{m_{p r o b e}} [v + ∆ v] - [v] = - v_{e} \ln (\frac{m_{p r o b e, f}}{m_{p r o b e, i}}) ∆ v = - v_{e} \ln (\frac{m_{p r o b e, f}}{m_{p r o b e, i}}) m_{p r o b e, f} = m_{p r o b e, i} e^{- (\frac{∆ v}{v_{e}})} (5) m_{p r o b e, f} = (800 lb) e^{- (\frac{∆ v}{v_{e}})}$

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