Orbital Speed calculation using conservation of Energy

[JayKo]

An unmanned spacecraft moves in a circular orbit with speed v around the moon,observing the lunar surface at an orbital radius of R=1790 km.The period of the orbit is T.Unfortunately, an electrical fault causes an on board thruster to fire, reducing the speed of the spacecraft by $$\Delta$$v=5 m/s. If nothing is done to correct its orbits. with what speed (in km/h) will the spacecraft crash into the lunar surface? Mass of moon=7.35$$\times$$10$$^{22}$$ kg and Radius of Moon = 1740 km. Universal gravitational constant, G = 6.673$$\times$$10$$^{-11}$$ Nm$$^{2}$$/kg$$^{2}$$.

Now the question is, after calculating using conservation of Energy Method,
[K+U]$$_{initial}$$=[K+U]$$_{final}$$.

the landing speed of the spacecraft is smaller than the escape velocity. why is that so? any one?

If the spacecraft orbiting without the electrical fault, is that speed suppose to be greater than,or lesser than then escape velocity in order for it to maintain at that orbit.

Thanks

 

[fleem]

Escape velocity is that velocity necessary to give a craft enough kinetic energy to completely leave the planet and enter interplanetary space. So by definition, a craft in orbit must have something less than this--otherwise it would escape.

Thus the impact velocity in the above story must be less than escape velocity, for if we watched time in reverse starting at the point of impact, we would see the craft return to orbit but not escape.

 

[JayKo]

hi Fleem, thanks for the prompt reply and clearing my doubts ;)
what you mean by saying
<if we watched time in reverse starting at the point of impact, we would see the craft return to orbit but not escape.>
I can understand if there is no electrical fault, the craft will maintain at it orbital speed and travel is circular orbitsince orbital speed is less than escape velocity, what force is exerted on the craft itself to counter the gravitation force of the moon from pulling it toward the moon?
is it due to Centrifugal force of circular motion by the craft? thanks

 

[fleem]

hi Fleem, thanks for the prompt reply and clearing my doubts ;)
what you mean by saying
<if we watched time in reverse starting at the point of impact, we would see the craft return to orbit but not escape.>
I can understand if there is no electrical fault, the craft will maintain at it orbital speed and travel is circular orbit

If you watched a backward film of something like this, orbital mechanics, conservation of energy, and so on would still be obeyed. I gave the example to help illustrate that escape velocity is also the impact velocity of an object dropped from an extreme (interplanetary) height. The process is reversible because of conservation of energy--converting potential energy to kinetic energy conserves just as well as converting kinetic energy to potential energy. Sorry if I added to the confusion.

since orbital speed is less than escape velocity, what force is exerted on the craft itself to counter the gravitation force of the moon from pulling it toward the moon?
is it due to Centrifugal force of circular motion by the craft? thanks

None. There is no force that counters the moon's force. The craft most certainly is falling toward the moon.

However, it was also given some tangential velocity so that as it falls toward the moon, it misses the moon before it crashes (assuming it is given enough tangential velocity). Specifically, that tangential velocity causes it to fall over the moon's horizon before it crashes. Once over the horizon, it finds itself in exactly the same situation: falling toward the moon but with a sufficient tangential velocity to make it miss the moon and fall over the horizon once more. the process is continuous as the craft travels around the moon--its always falling toward the moon but just missing it as it moves over the horizon because of that tangential velocity.

 

[JayKo]

If you watched a backward film of something like this, orbital mechanics, conservation of energy, and so on would still be obeyed. I gave the example to help illustrate that escape velocity is also the impact velocity of an object dropped from an extreme (interplanetary) height. The process is reversible because of conservation of energy--converting potential energy to kinetic energy conserves just as well as converting kinetic energy to potential energy. Sorry if I added to the confusion.



None. There is no force that counters the moon's force. The craft most certainly is falling toward the moon.

However, it was also given some tangential velocity so that as it falls toward the moon, it misses the moon before it crashes (assuming it is given enough tangential velocity). Specifically, that tangential velocity causes it to fall over the moon's horizon before it crashes. Once over the horizon, it finds itself in exactly the same situation: falling toward the moon but with a sufficient tangential velocity to make it miss the moon and fall over the horizon once more. the process is continuous as the craft travels around the moon--its always falling toward the moon but just missing it as it moves over the horizon because of that tangential velocity.

Hi fleem,

Thanks for the detail explanation,will go through it,and if any question will post it later,mean while ,really appreciate your time in this post ;)

 

[JayKo]

If you watched a backward film of something like this, orbital mechanics, conservation of energy, and so on would still be obeyed. I gave the example to help illustrate that escape velocity is also the impact velocity of an object dropped from an extreme (interplanetary) height. The process is reversible because of conservation of energy--converting potential energy to kinetic energy conserves just as well as converting kinetic energy to potential energy. Sorry if I added to the confusion.

not a problem,you already cleared up the confusion. i understand the part of inter conversation of potential energy and kinetic energy and vice versa as the orbital radius changes ;) thanks.

None. There is no force that counters the moon's force. The craft most certainly is falling toward the moon.

However, it was also given some tangential velocity so that as it falls toward the moon, it misses the moon before it crashes (assuming it is given enough tangential velocity). Specifically, that tangential velocity causes it to fall over the moon's horizon before it crashes. Once over the horizon, it finds itself in exactly the same situation: falling toward the moon but with a sufficient tangential velocity to make it miss the moon and fall over the horizon once more. the process is continuous as the craft travels around the moon--its always falling toward the moon but just missing it as it moves over the horizon because of that tangential velocity.

By the law of gravitational, the net force,$$^{}$$
F=$$\frac{Gm_{m}m}{r^{2}}$$
And by Netwon 2nd laws,$$\Sigma$$$$\stackrel{\rightarrow}{F}$$=m$$\stackrel{\rightarrow}{a}$$=$$\frac{mv^{2}}{r}$$
Equating both forces, we get
$$\frac{Gm_{m}m}{r^{2}}$$=$$\frac{mv^{2}}{r}$$
hence, tangential velocity,v=$$\sqrt{\frac{Gm_{m}}{r}}$$
where $$m_{m}}$$ is the mass of the moon. this is the tangential velocity that you mention above.right? because of the tangential velocity component it gives the satellite a circular orbit,without crashing into the moon. this is how i interpret your explanation above, right?

thanks ;)

 

[JayKo]

now we come to final part of the question.

If the scientist on Earth manages to veer the spacecraft onto a circular orbit of slightly larger radius R+$$\Delta$$R, where $$\Delta$$R<<R, such that the speed of the spacecraft in this orbit is v-$$\Delta$$v, determine a relation between the quantities $$\Delta$$v,$$\Delta$$R and T.Noted that $$\Delta$$R,$$\Delta$$v are positive quantities.
(hint: use the expression$$(1+x)^{n}$$ $$\approx$$ 1+nx,valid for |x|<<1).how should i first go about it, kindly advise.thanks. ;)
what i can think of is using the conversation of energy,and plug in the variable into the equation.is that the right direction to start?

 

[tony873004]

Why do you think a delta V of only 5 m/s would make the spaceship crash to the surface, rather than simply lower its perilune?

 

[JayKo]

first, we find the speed require for the spacecraft to orbit the moon using $$\sqrt{\frac{Gm_{m}}{r}}$$.which is 1655.30 m/s

Due to the -5m/s speed change, the orbit speed became 1650.3 m/s
Plug in 1650.3 as initial speed in [K+U]before=[K+U]after.
we get the impact speed = 1697.34m/s
I m using conversation of energy model to solve this problem,not sure if there is any other approach in minds?

the explanation is:
after the electrical fault, the spacecraft is moving too slowly to be in a stable orbit; the gravitational force is larger that what is needed to maintain a circular orbit. the spacecraft gains energy as it is accelerated toward the surface.

 

[tony873004]

Your formula for orbital velocity is wrong. You need to add the 2 m's instead of multiply them: $$\sqrt{\frac{G(m+_{m})}{r}}$$ . But you got the right answer, so it's probably just a typo.

$$\begin{array}{l}<br /> r_{{\rm{perilune}}} = \frac{{r_{{\rm{apolune}}} }}{{\left( {\frac{{2GM}}{{r_{{\rm{apolune}}} v_{{\rm{apolune}}}^2 }} - 1} \right)}} = \frac{{1790000m}}{{\left( {\frac{{2 \cdot 6.6725985 \times 10^{ - 11} m^3 kg^{ - 1} s^{ - 2} \cdot 7.34775349 \times 10^{22} kg}}{{1790000m \cdot \left( {1650m/s} \right)^2 }} - 1} \right)}} = 1769000m = 1769km \\ <br /> \\ <br /> {\rm{altitude}} = r_{{\rm{perilune}}} - r_{{\rm{Moon}}} = 1769km - 1737km = 32km \\ <br /> \end{array}$$

You're closest approach to the Moon is 32 kilometers above its surface. This question is flawed. You're not going to hit the Moon. You're simply transferred to an elliptical orbit. You need to lose another 8 m/s for a transfer orbit that takes you to the surface.

 

[JayKo]

Your formula for orbital velocity is wrong. You need to add the 2 m's instead of multiply them: $$\sqrt{\frac{G(m+_{m})}{r}}$$ . But you got the right answer, so it's probably just a typo.

$$\begin{array}{l}<br /> r_{{\rm{perilune}}} = \frac{{r_{{\rm{apolune}}} }}{{\left( {\frac{{2GM}}{{r_{{\rm{apolune}}} v_{{\rm{apolune}}}^2 }} - 1} \right)}} = \frac{{1790000m}}{{\left( {\frac{{2 \cdot 6.6725985 \times 10^{ - 11} m^3 kg^{ - 1} s^{ - 2} \cdot 7.34775349 \times 10^{22} kg}}{{1790000m \cdot \left( {1650m/s} \right)^2 }} - 1} \right)}} = 1769000m = 1769km \\ <br /> \\ <br /> {\rm{altitude}} = r_{{\rm{perilune}}} - r_{{\rm{Moon}}} = 1769km - 1737km = 32km \\ <br /> \end{array}$$

You're closest approach to the Moon is 32 kilometers above its surface. This question is flawed. You're not going to hit the Moon. You're simply transferred to an elliptical orbit. You need to lose another 8 m/s for a transfer orbit that takes you to the surface.

Hi Tony, thanks for the detail solution, the small m is the subscript of the big m,$$\sqrt{\frac{G(m_{m})}{r}}$$.(which is derived at post #6) which means mass of the moon.

By the way, the velocity of the satellite doesn't depend on its own mass,but the mass of the moon,correct me if i am wrong.

your equation of finding radius of perilune is new to me, this is a undergraduate physics question. nonetheless this is interesting discussion going on here. i try to Wiki perilune, and get to know more about this celestial vocabulary. But what is the concept behind the formula that you used,it don't look like Kepler to me;)

 

[tony873004]

Ok, as long as Mm means Mass of the Moon and not Mass of Moon * mass of spacecraft , you're ok, since the mass of the spacecraft is negligable here.

For more on that formula, see: http://www.braeunig.us/space/
It's equation 3.19. Robert Braeunig had done a really nice job of collecting lots of equations relavant to orbital mechanics on this web page, along with describing what the formulas mean.

As a double check, I like to simulate these things with Gravity Simulator. The simulated result and the result from formula 3.19 are in agreement with each other.

I didn't double-check your energy equations, but your answer looks about what I'd expect it to be if there were an impact. So in addition to solving it the way your teacher wants, turn in the real answer too for some extra credit.

 

[JayKo]

Ok, as long as Mm means Mass of the Moon and not Mass of Moon * mass of spacecraft , you're ok, since the mass of the spacecraft is negligable here.

For more on that formula, see: http://www.braeunig.us/space/
It's equation 3.19. Robert Braeunig had done a really nice job of collecting lots of equations relavant to orbital mechanics on this web page, along with describing what the formulas mean.

As a double check, I like to simulate these things with Gravity Simulator. The simulated result and the result from formula 3.19 are in agreement with each other.

I didn't double-check your energy equations, but your answer looks about what I'd expect it to be if there were an impact. So in addition to solving it the way your teacher wants, turn in the real answer too for some extra credit.

yupe Mm is just mass of moon. well thanks again. by the way you analyze this question in a very professional way, just wondering if you are working for NASA? i go check out the website now ;)

 

[JayKo]

Ok, as long as Mm means Mass of the Moon and not Mass of Moon * mass of spacecraft , you're ok, since the mass of the spacecraft is negligable here.

For more on that formula, see: http://www.braeunig.us/space/
It's equation 3.19. Robert Braeunig had done a really nice job of collecting lots of equations relavant to orbital mechanics on this web page, along with describing what the formulas mean.

As a double check, I like to simulate these things with Gravity Simulator. The simulated result and the result from formula 3.19 are in agreement with each other.

I didn't double-check your energy equations, but your answer looks about what I'd expect it to be if there were an impact. So in addition to solving it the way your teacher wants, turn in the real answer too for some extra credit.

Hi Tony,I did check at the equation which is actually derived from Conservation of Energy, instead of making velocity the subject, Eq 3.19 making the radius as subject, so both are derived from the same energy concept.I would have to re-calculate again.

 

[tony873004]

...just wondering if you are working for NASA?

not yet :). I'm an undrad majoring in Physics / Astronomy.

Let me know what your teacher thinks of your answer. I'm sure it will be more detailed than the other student's answers.

 

[JayKo]

not yet :). I'm an undrad majoring in Physics / Astronomy.

Let me know what your teacher thinks of your answer. I'm sure it will be more detailed than the other student's answers.

oh i see, well i m afraid my professor will not be able comment until this paper is mark and the result is released,this was our exam question ;)