Is this solution correct? (amount of fuel needed for orbital maneuvers)

In summary: I think that's the right word. The idea that one might want to know the energy content of a fuel is somewhat distinct from the idea that one might want to know the mass of fuel needed to accomplish some task. Combining them is not only a bit confusing, but also makes the whole thing an exercise in dimensional analysis. Which is fine, but not very interesting. And not at all relevant to the task of describing the energy content of the satellite's orbit.But I'm not an expert in summarizing content, so I'll stop here.
  • #1
Nirmal Padwal
41
2
Homework Statement
Expressing kinetic energy of a satellite in a circular orbit in terms of the height of the satellite above the earth's surface (##h##), calculate the amount of hydrogen (energy density: ##10^6## J/Litre) needed to bring a small 1.5 kg satellite in an orbit of 200 km. (Given: ##m_e## = 5.972##\times 10^{24}## kg, ##R_e## = 6371 km = 6371##\times 10^3## m, ##G## = 6.67##\times 10^{-11}## N.m##^2##.kg##^{-2}##
Relevant Equations
1. $$K.E. = G \frac{m_s . m_e}{2(R_e+h)}\\$$
2. $$ \mathrm{Required \ amount} = \frac{E_{kin}(h)}{E_{dens}}$$
The kinetic energy of a ##m_s##= 1.5 kg satellite in an orbit of ##h## = 200 km = 200##\times 10^3## m is
\begin{eqnarray*}
K.E. &=& G \frac{m_s . m_e}{2(R_e+h)}\\
&=& (6.67 \times 10^{-11} \ \mathrm{N.m^2.kg^{-2}}) \times
\frac{(1.5\ \mathrm{kg})\times (5.972\times 10^{24}\ \mathrm{kg})}{2\times (6371+200) \times 10^3
\ \mathrm{m}}\\
E_{kin}(h) &=& 45.465 \times 10^6\ \mathrm{J}
\end{eqnarray*}

As energy density of liquid hydrogen is given to be ##E_{dens}## = ##10^6## J/Litre, total amount of liquid hydrogen required for the satellite to gain kinetic energy of 45.465 ##\times 10^6## J is
$$ \mathrm{Required \ amount} = \frac{E_{kin}(h)}{E_{dens}} = \frac{45.465 \times 10^6\ \mathrm{J}}{10^6 \ \mathrm{J/Litre}}= 45.465 \ \mathrm{Litre}$$

I am not sure if this solution is right. Please let me know if something is not right.
 
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  • #2
You use the virial theorem correctly for your ##E_{k}## expression, however, mightn't you also need to take into account the change in potential energy? That is to say, if we imagine the Earth is not rotating, the satellite is initially at rest on the surface and at the end it's at a new height with some kinetic energy. The total external work done by the exhaust stream on the rocket must account for both of these changes.

You know that the total energy of the orbit is ##E=-E_{k} = \frac{1}{2}U##.
 
  • #3
Yes, I should ideally take that into account. But they haven't provided any information about the rocket (like the initial mass of the rocket, exhaust velocity, etc.)

But is my solution correct?
 
  • #4
Nirmal Padwal said:
Yes, I should ideally take that into account. But they haven't provided any information about the rocket (like the initial mass of the rocket, exhaust velocity, etc.)

But is my solution correct?

You can ignore the fuel and variable mass for this question. But the total energy requirement is going to also include that needed to change the potential energy of the Earth-rocket system. So I wouldn't say your answer is correct yet.
 
  • #5
I think I understood your point. The change in the potential energy as the rocket moves from ##h = 0 ## km to ##h = 200 ## km is $$U = G.m_s.m_e. \left( \frac{1}{R_e+200} - \frac{1}{R_e}\right) $$ Substituting the respective values, I got ##U= -2.845 \times 10^{6}## J

So the total energy is ##(45.465 - 2.845)\times 10^{6} = 42.62 \times 10^{6}## J and the and the required amount of fuel is ##42.62## Litre. Is this correct?
 
  • #6
Nirmal Padwal said:
The change in the potential energy as the rocket moves from ##h = 0 ## km to ##h = 200 ## km is $$\mathbf{\Delta} U = G.m_s.m_e. \left( \frac{1}{R_e+200} - \frac{1}{R_e}\right) $$

Are you sure the signs are right? Does the potential energy decrease if you increase the separation between the masses :wink:?
 
  • #7
Oops.. it increases:sorry:. So the total energy will be ##(45.465 + 2.845)\times 10^{6} = 48.31 \times 10^{6}## J
 
  • #8
Looks good! Of course, to get a better estimate you'd need to account for losses to drag, the initial kinetic energy provided by the Earth's rotation and all sorts!
 
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  • #9
Thank you so much for your help!
 
  • #10
Nirmal Padwal said:
Homework Statement:: Expressing kinetic energy of a satellite in a circular orbit in terms of the height of the satellite above the Earth's surface (##h##), calculate the amount of hydrogen (energy density: ##10^6## J/Litre) needed to bring a small 1.5 kg satellite in an orbit of 200 km. (Given: ##m_e## = 5.972##\times 10^{24}## kg, ##R_e## = 6371 km = 6371##\times 10^3## m, ##G## = 6.67##\times 10^{-11}## N.m##^2##.kg##^{-2}##
Relevant Equations:: 1. $$K.E. = G \frac{m_s . m_e}{2(R_e+h)}\\$$
2. $$ \mathrm{Required \ amount} = \frac{E_{kin}(h)}{E_{dens}}$$
I am not sure if this solution is right. Please let me know if something is not right.
These equations assume the truth of two questionable assumptions.

1. When you burn rocket fuel and expel it from the back of a rocket, does 100% of the energy go into the forward motion of the rocket?

2. If you get to the end of the calculation, can you compare the mass of the propellant used to the mass of the payload? Does the solution account for the energy expended on the fuel in the tank during the burn?

Tsiolkovsky has an equation for this. But one would need to know the fuel's energy density per unit mass.

I consider this a poorly posed problem. It invites you to make assumptions so poor that the resulting answer is not even close to the real world answer. It does not give you the information that would be required to compute the correct answer.

Edit: The energy density of hydrogen is zero. Without an oxidizer, it does not burn.
 
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  • #11
Since no details on the exhaust stream are given, the question is essentially abstracting the scenario to a body of constant mass being pulled by some arbitrary external force. This is not necessarily a bad thing, since otherwise the question would just gain added layers of complexity,

But the fact that it then asks you to work out another number pertaining to the fuel is misleading, since it's now conflating two different models.

And yes.. the answer is slightly ridiculous :wink:... you could buy that much milk for quite less than a fifty...
 
  • #12
But the answer to the first part is very interesting. The energy cost of putting 1 kg into low Earth orbit is 30 Mjoules=10 kWhr. So the problem is not one of Physics but that our technology is really lousy for putting things into space.
Not a very good problem but a very useful exercise.
 
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  • #13
hutchphd said:
But the answer to the first part is very interesting. The energy cost of putting 1 kg into low Earth orbit is 30 Mjoules=10 kWhr. So the problem is not one of Physics but that our technology is really lousy for putting things into space.
Not a very good problem but a very useful exercise.

Makes one of these space elevators look like a good idea:

1589112872462.png
 
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Related to Is this solution correct? (amount of fuel needed for orbital maneuvers)

1. How do you calculate the amount of fuel needed for orbital maneuvers?

The amount of fuel needed for orbital maneuvers is calculated using the rocket equation, which takes into account the mass of the spacecraft, the desired change in velocity, and the specific impulse of the rocket engine. This equation is used to determine the delta-v (change in velocity) required for the maneuver, and then the amount of fuel needed is calculated based on the specific impulse of the engine.

2. What factors affect the amount of fuel needed for orbital maneuvers?

The amount of fuel needed for orbital maneuvers can be affected by several factors, including the mass of the spacecraft, the desired change in velocity, the specific impulse of the rocket engine, and the efficiency of the engine. Other factors such as atmospheric drag and gravitational forces may also play a role.

3. How accurate are the calculations for the amount of fuel needed for orbital maneuvers?

The accuracy of the calculations for the amount of fuel needed for orbital maneuvers depends on the accuracy of the input data and the assumptions made in the calculations. In real-world scenarios, there may be variations and uncertainties that can affect the accuracy of the calculations.

4. Can the amount of fuel needed for orbital maneuvers be reduced?

Yes, the amount of fuel needed for orbital maneuvers can be reduced by optimizing the spacecraft design and using more efficient rocket engines. Additionally, advanced techniques such as aerobraking and gravity assists can also help reduce the amount of fuel needed for orbital maneuvers.

5. Are there any alternative methods for determining the amount of fuel needed for orbital maneuvers?

Yes, there are alternative methods for determining the amount of fuel needed for orbital maneuvers, such as computer simulations and empirical data from previous missions. However, these methods may not always be as accurate as using the rocket equation and may require additional adjustments and considerations.

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