A rocket traveling 1950 m/s away from the Earth at an altitude...

Question

A rocket traveling 1950 m/s away from the Earth at an altitude of 6400 km fires its rockets, which eject gas at a speed of 1200 m/s (relative to the rocket). If the mass of the rocket at this moment is 25,000 kg and an acceleration of 1.5 m/s² is desired, at what rate must the gases be ejected?

Accepted Answer

Low. Let's go through this practice problem. What is the necessary rate of propellant expulsion for a spacecraft moving at 1350 m per second away from earth at an altitude of 6500 kilometers to activate its thrusters, expelling propellant at a speed of 1210 m per second relative to the spacecraft, given the spacecraft's mass at this moment is 25,500 kg and desires an acceleration of 1.6 m per second squared option. A negative 36 kg per second. B 67 kg per second. C negative 76 kg per second or d 83 kg per second. So let's think about what this problem is asking. So let's say this is the earth and we've got a rocket that is flying away from us. In order for the rocket to fly, it needs to expel mass this propellant away from it. In order to keep the rocket moving. This problem is asking us what it for the minimum rate of propellant expulsion. So in other words, the minimum mass loss rate, in order for the rocket to overcome any external forces that might be holding it back. So in other words, we're looking at the case where the net external forces acting on the rocket is equal to the forces that the rocket itself is creating to push it away from the earth. So this is gonna be equal to the mass multiplied by the acceleration of the rocket or the rate of change of its speed minus the rate of change of mass due to the fact that we are losing mass when, as we are expelling the propellant. So minus the relative speed, we'll call it V sub RL, the relative speed of the propellant multiplied by the rate of change of mass. And that rate of change of mass is what we're trying to solve in this problem. We also need to consider those external forces before we can continue. So let's say that the mass of earth has a massive M sub earth. I'll just call it M sub E. The mass of the rocket will just say is a mass of M. So the only external force that should be acting on this rocket is the gravitational force between the rocket and the earth. So those external forces are equal to nothing but the gravitational force which is equal to the constant big G multiplied by the mass of the earth multiplied by M divided by R squared. Since this force is going to cause the rocket to be tended to accelerate towards the earth. In other words, away from its direction, I'll label it negative when you plug it back into the force equation, that's negative GM sub em divided by R squared equals M DVD T minus the row D MDT. So now let's rewrite this equation to solve it for D MDT, which is what the problem is asking for. First off, let's get the term containing the D MDT on its own or we can do that by adding both sides of the equation V Sorel D MDT. And then also moving the, the gravitational force term over to the right hand side of the equation as well. We have V round D MDT equals MD V DT plus G MC M divided by R squared. You can also simplify the right hand side of this slightly by factoring out the M. So M multiplied by DVD T plus GM sub E divided by R squared. Then we'll just divide both sides of the equation by V relative. So D MDT is equal to M divided by VR multiplied by DVD T plus GM sub E divided by R squared. So all that's left for us to do is plug into this equation. The variables that were given to us in the problem. So the mass of the rocket is given as 25,500 kg. We divide this by the relative speed of the fuel which is given as negative 1350 m per second. We multiply it by DVD T which is just this acceleration required, which is 1.6 m per second squared plus big G, which is 6.67 multiplied by 10 to the power of a negative 11 Newton meters squared per kilogram squared multiplied by the mass of the earth which is 5.98 multiplied by 10. The power of 24 kg divided by R squared. That's the square of the distance between the rocket and the earth's center of mass. So that means we need to add the radius of the earth 6.38 multiplied by 10 to the power of six meters plus the altitude which is 6500 kilometers or 6.5 multiplied by 10 to the power of 6 m square that and put this all into a calculator. And we find a mass rate change of about negative 75.6 kilograms per second, which can be rounded to about negative 76.0 kilograms per second. And so that is our answer to this problem which corresponds to option C. So option C is our answer now. That's it for this problem. And I hope this video helped you out. If it did, please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

A rocket traveling 1950 m/s away from the Earth at an altitude of 6400 km fires its rockets, which eject gas at a speed of 1200 m/s (relative to the rocket). If the mass of the rocket at this moment is 25,000 kg and an acceleration of 1.5 m/s² is desired, at what rate must the gases be ejected?

Key Concepts

Rocket Propulsion

Thrust and Acceleration

Mass Flow Rate

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