Calculate the time for a spacecraft to arrive at Mars (No Kepler's law)

[jisbon]

Homework Statement

A minimum-energy transfer orbit to an outer planet consists of putting a spacecraft on
an elliptical trajectory with the departure planet corresponding to the perihelion of the ellipse, or the closest point to the Sun, and the arrival planet at the aphelion, or the farthest point from the Sun. (Assume the orbital radius of the Earth is $1.50*10^8$ m, and the orbital radius of Mars is $2.30*10^8$ m, and mass of Sun to be $2.00*10^30$kg. Calculate how long spacecraft will take to arrive at Mars.

Relevant Equations

NIL

So I've seen many similar questions like this on the internet, but they all involve using Kepler's law, which is not in my syllabus in my case. However, they do give me the mass of the Sun. which I assume involves gravitational force/field/circular motion?
The question also gave me a hint that I could work out the formula by assuming radius r and then replace with semi-major axis a. Not sure what it is talking about exactly.
Sorry, but I don't really know how to start this, but advice will be appreciated. Thanks!

 

[haruspex]

Homework Statement: A minimum-energy transfer orbit to an outer planet consists of putting a spacecraft on
an elliptical trajectory with the departure planet corresponding to the perihelion of the ellipse, or the closest point to the Sun, and the arrival planet at the aphelion, or the farthest point from the Sun. (Assume the orbital radius of the Earth is $1.50*10^8$ m, and the orbital radius of Mars is $2.30*10^8$ m, and mass of Sun to be $2.00*10^30$kg. Calculate how long spacecraft will take to arrive at Mars.
Homework Equations: NIL

View attachment 250073

So I've seen many similar questions like this on the internet, but they all involve using Kepler's law, which is not in my syllabus in my case. However, they do give me the mass of the Sun. which I assume involves gravitational force/field/circular motion?
The question also gave me a hint that I could work out the formula by assuming radius r and then replace with semi-major axis a. Not sure what it is talking about exactly.
Sorry, but I don't really know how to start this, but advice will be appreciated. Thanks!

Can you start by computing the semimajor axis?

 

[jisbon]

Can you start by computing the semimajor axis?

I'm not sure where is the semi major axis/what is it even. Any clues to that? Thanks

 

[haruspex]

I'm not sure where is the semi major axis/what is it even. Any clues to that? Thanks

https://en.m.wikipedia.org/wiki/Semi-major_and_semi-minor_axes

 

[jisbon]

https://en.m.wikipedia.org/wiki/Semi-major_and_semi-minor_axes

OK, so that is like a ellipse. So to compute semi major axis, isn't it just the radius since we can assume semi major axis is half the long axis of the ellipse? (AKA then diameter)? What do I do next after knowing the semi major axis?

 

[haruspex]

OK, so that is like a ellipse. So to compute semi major axis, isn't it just the radius since we can assume semi major axis is half the long axis of the ellipse? (AKA then diameter)? What do I do next after knowing the semi major axis?

You can assume the launch is timed such that Mars will be in the right place at the time its orbit is reached. So you just need the time to go around half the ellipse.
The hint allows you to pretend instead that it goes around half a circle.
How long does the Earth take to do that? What about Mars?

 

[Delta2]

OK, so that is like a ellipse. So to compute semi major axis, isn't it just the radius since we can assume semi major axis is half the long axis of the ellipse? (AKA then diameter)? What do I do next after knowing the semi major axis?

I think the scheme that you posted is very good and it allow us to determine the major axis of the ellipse by just looking at it. I think the major axis of the elliptical orbit of the spacecraft is $r_{Mars}+r_{Earth}$.

 

[jisbon]

I think the scheme that you posted is very good and it allow us to determine the major axis of the ellipse by just looking at it. I think the major axis of the elliptical orbit of the spacecraft is $r_{Mars}+r_{Earth}$.

Yep, I can see that the major axis of the elliptical orbit of the spacecraft is $r_{Mars}+r_{Earth}$. With this information of the distance of the major axis, what will I do next? Incorporate gravitational force between the sun and the ship or? And how do I exactly get a period out of the force?
Thanks

 

[Delta2]

Yep, I can see that the major axis of the elliptical orbit of the spacecraft is $r_{Mars}+r_{Earth}$. With this information of the distance of the major axis, what will I do next? Incorporate gravitational force between the sun and the ship or? And how do I exactly get a period out of the force?
Thanks

Can you post exactly the hint given by your book (that about assuming radius r and replacing with the semi major axis). I think your book means that the elliptical orbit of the spaceship is approximately equivalent to a circular orbit of radius $r=\frac{r_{Mars}+r_{Earth}}{2}$. And all you got to do is finding the semi period of this circular orbit around the sun.

 

[jisbon]

Can you post exactly the hint given by your book (that about assuming radius r and replacing with the semi major axis). I think your book means that the elliptical orbit of the spaceship is approximately equivalent to a circular orbit of radius $r=\frac{r_{Mars}+r_{Earth}}{2}$. And all you got to do is finding the semi period of this circular orbit around the sun.

The hint is exactly as follows:

You can work out the required formula by assuming circular radius r, then replace with the semi-major axis a. The semi-major axis is half the long axis of the cliipise.

And by finding the semi period will this be correct?

$\frac{GMm}{r^2}=\frac{mv^2}{r}$
$v =\sqrt{\frac{GM}{r}}$
Orbital speed = $\frac{Distance travelled}{T}$
where Distance = $\frac{r_{Mars}+r_{Earth}}{2}$ ?
$T = Distance / speed$
$= \frac{r_{Mars}+r_{Earth}}{2} / \sqrt{\frac{GM}{r}}$
?
Thanks

 

[Delta2]

Almost correct, but distance traveled is equal to $\frac{2\pi\frac{r_{Mars}+r_{Earth}}{2}}{2}$, that is the distance of a half circle of radius $\frac{r_{Mars}+r_{Earth}}{2}$.

 

[jisbon]

Almost correct, but distance traveled is equal to $\frac{2\pi\frac{r_{Mars}+r_{Earth}}{2}}{2}$, that is the distance of a half circle of radius $\frac{r_{Mars}+r_{Earth}}{2}$.

Yep, forgot the formula for the radius. Will get back to you to see if I can get it correct this time

EDIT: Seems to be wrong. Did I make any mistakes along the way?

$\frac{\frac{\pi \left(\left(2.3\cdot 10^8\right)+\left(1.5\cdot 10^8\right)\right)}{2}}{\sqrt{\frac{\left(6.673\cdot \:10^{-11}\right)\left(2\cdot 10^{30}\right)}{\frac{\left(2.3\cdot \:10^8\right)+\left(1.5\cdot \:10^8\right)}{2}}}}$ = 712...

 

[haruspex]

Yep, forgot the formula for the radius. Will get back to you to see if I can get it correct this time

EDIT: Seems to be wrong. Did I make any mistakes along the way?

$\frac{\frac{\pi \left(\left(2.3\cdot 10^8\right)+\left(1.5\cdot 10^8\right)\right)}{2}}{\sqrt{\frac{\left(6.673\cdot \:10^{-11}\right)\left(2\cdot 10^{30}\right)}{\frac{\left(2.3\cdot \:10^8\right)+\left(1.5\cdot \:10^8\right)}{2}}}}$ = 712...

THe period is proportional to r³, right? So you can quickly derive it as a ratio from Earths's period. I get 742 days.

 

[Delta2]

THe period is proportional to r³, right? So you can quickly derive it as a ratio from Earths's period. I get 742 days.

You mean the square of the period is proportional to $r^3$. That's kepler's 3rd law. So does this give 742 days or is 742 days a blunder assuming the period is proportional to $r^3$?

 

[Delta2]

I get around 520 days period, 260 days the time for the spacecraft since it is half period.

 

[jisbon]

You mean the square of the period is proportional to $r^3$. That's kepler's 3rd law. So does this give 742 days or is 742 days a blunder assuming the period is proportional to $r^3$?

I don't know if I'm supposed to use keplers law in this scenario because I have not been taught about it. I was thinking more along the lines of using gravitational forces to solve for this

 

[Delta2]

Essentially your work is a proof of Kepler's 3rd law... you end up to the result that
$$T=\frac{2\pi r}{\sqrt{\frac{GM_{sun}}{r}}}\Rightarrow T^2=\frac{4\pi^2}{GM_{sun}}r^3$$

EDIT: I just figured out what's wrong with your calculations: There is a mistake in the units of the given orbital radius of Earth and mars, the units should be in Km not in meters.

 

[jisbon]

Essentially your work is a proof of Kepler's 3rd law... you end up to the result that
$$T=\frac{2\pi r}{\sqrt{\frac{GM_{sun}}{r}}}\Rightarrow T^2=\frac{4\pi^2}{GM_{sun}}r^3$$

EDIT: I just figured out what's wrong with your calculations: There is a mistake in the units of the given orbital radius of Earth and mars, the units should be in Km not in meters.

Yes, I realized I mistyped the question statement. It should be in KM, so values should still hold.

The answer given is 0.714 years, which is still wrong even if it's problems with the units. Any ideas?

 

[Delta2]

0.714 years (earthen years) is about 260 days just as I said in post #15

Check again your calculations, the LHS of the equation in post #12 is correct, it just that where you have $10^8$ you should put $10^{11}$

 

[jisbon]

0.714 years (earthen years) is about 260 days just as I said in post #15

Check again your calculations, the LHS of the equation in post #12 is correct, it just that where you have $10^8$ you should put $10^{11}$

Got it! Guess I have to convert the km to m first (SI Units probably). Thanks!