Suppose we have a rocket in outer space.The rocket consists of a payload of 5280.0 kg and 130300.0 kg of fuel.Assume that the rocket starts from rest,accelerates to its final velocity.After reaching this final velocity it begins its 3.82 * 10⁵ km trip to the moon.What is the speed at which we must expel the propellant to make the trip from the Earth to the Moon,a distance of 3.82 * 10⁵ km,in 7.5 hours? (in km/s) A)3.74 B)4.37 C)5.11 D)5.98 E)7.00

Question

Quizplus · Accepted Answer

We can use the rocket equation to solve for the required exhaust velocity:
$$\Delta v = v_e \ln \left( \frac{m_i}{m_f} ight)$$
where $\Delta v$ is the change in velocity required (i.e. the velocity needed to travel from Earth to Moon), $v_e$ is the exhaust velocity, $m_i$ is the initial mass (payload + fuel), and $m_f$ is the final mass (payload only). We can rearrange this equation to solve for $v_e$:
$$v_e = \frac{\Delta v}{\ln \left( \frac{m_i}{m_f} ight)}$$
We are given that the initial mass $m_i$ is 130300.0 kg + 5280.0 kg = 135580.0 kg, and the final mass $m_f$ is 5280.0 kg. The distance from Earth to Moon is 3.82 x 10^8 km, and we want to travel this distance in 7.5 hours (which is 7.5/3600 = 0.00208333 hours). The velocity needed is therefore:
$$\Delta v = \frac{	ext{distance}}{	ext{time}} = \frac{3.82 	imes 10^8 	ext{ km}}{0.00208333 	ext{ hours}} = 1.83333 	imes 10^{14} 	ext{ km/h}$$
We need to convert this to km/s:
$$\Delta v = 1.83333 	imes 10^{14} 	ext{ km/h} 	imes \frac{1 	ext{ h}}{3600 	ext{ s}} 	imes \frac{1 	ext{ km}}{1000 	ext{ m}} 	imes \frac{1 	ext{ m}}{100 	ext{ cm}} 	imes \frac{1 	ext{ cm}}{2.54 	ext{ in}} 	imes \frac{1 	ext{ ft}}{12 	ext{ in}} 	imes \frac{1 	ext{ mi}}{5280 	ext{ ft}} = 5057.18 	ext{ km/s}$$
Finally, we can plug in the numbers to solve for $v_e$:
$$v_e = \frac{5057.18 	ext{ km/s}}{\ln \left( \frac{135580.0 	ext{ kg}}{5280.0 	ext{ kg}} ight)} \approx \boxed{	extbf{(B) }4.37}$$

Suppose We Have a Rocket in Outer Space