Question

A 70-kg astronaut floating in space in a 110-kg MMU (manned maneuvering unit) experiences an acceleration of 0.029 m/s2 when he fires one of the MMU's thrusters. You may want to review (Pages 258 - 260) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Acceleration of a rocket. If the speed of the escaping N2 gas relative to the astronaut is 490 m/s, how much gas is used by the thruster in 5.0s and what is the thrust of the thruster?

Answer 1

Answers:

a) 0.053 kg

b) 5.22 N

Explanation:

We have the following data:

$m_{A}= 70 kg$ is the mass of the astronaut

$M= 110 kg$ is the mass of the MMU

$a=0.029 m/s^{2}$ is the acceleration of the system astronaut-MMU

$V_{N_2}=490 m/s$ is the velocity of the ejected fuel

Now, we have the following two equations to find the amount of gas (mass)$dm$ ejected by the thruster and the thrust $F$:

$a=-\frac{V_{N_2}}{m} \frac{dm}{dt}$ (1)

$F=-V_{N_2} \frac{dm}{dt}$ (2)

Where:

$m=m_{A} + M=70 kg + 110 kg=180 kg$ is the total mass of the system astronaut-MMU

$dm$ is the mass of the ejected fuel

$dt=5 s$ is the time interval in which this occurs

Knowing this, let's begin with the answers:

a) How much gas is used by the thruster in 5s?

Here we have to isolate $dm$ from (1) considering $m$ as constant, since the mass of the fuel is negligible compared to the total mass $m$:

$dm=-\frac{m a}{V_{N_2}} dt$ (3)

$dm=-\frac{(180 kg)(0.029 m/s^{2})}{490 m/s} 5 s$ (4)

$dm=-0.053 kg$ (5) Note the negative sign is because the direction of the ejected mass is opposite to the direction of the system astronaut-MMU. However, the value of $dm$ is 0.053 kg.

b) What is the thrust of the thruster?

Let's use equation (2) with the $dm$ calculated in (5):

$F=(-490 m/s) \frac{-0.053 kg}{5s}$ (6)

Finally:

$F=5.22 N$ (7)