# Homework Help: Hohmann transfer towards mars: True Anomaly of Earth/ Launch date

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1. Nov 3, 2013

### Paul Gray

Hello again :),

I have to solve a rather extensive task. Most important are questions a) and c). I appreciate any help :)!

1. The problem statement, all variables and given/known data
The task: Determine characteristics of a Hohmann transfre towards, when circular planetary orbits around the Sun with $R_E = 149.6 \cdot 10^6 km$ for Earth and $R_M = 227.9 \cdot 10^6 km$ for Mars are used.

a) How does the true anomaly in [deg] of the Earth's position between start and arrival of the satellite change?
b) What is the angle [deg] between Earth / Sun/ Mars when the satellite arrives at Mars?
c) How many days before opposition (Earth is on a line between Sun and Mars) the launch from Earth must occur to arrive at the apocentre of the transfer ellipse, when Mars is to be encountered.

2. Relevant equations

The orbit transfer time $t_H = 258.8 [days]$

The orbital velocity of Earth $\nu_E = \sqrt{\frac{\mu_{Sun}}{R_E}} = 107224.1113 \frac{km}{h}$
The orbital velocity of Mars $\nu_M = \sqrt{\frac{\mu_{Sun}}{R_M}} = 86870 \frac{km}{h}$

The angular velocity of Earth $\omega_E = \frac{360^\circ}{365[days]} = 0.9863\frac{deg}{days}$
The angular velocity of Mars $\omega_M = \frac{360^\circ}{686.795[days]} = 0.5242\frac{deg}{days}$ (received from the orbital period T).

3. The attempt at a solution
a) For the true anomaly I only found a formula covering the "One-tangent burn" orbit changing technique: $\nu = \arccos{(\frac{a_{tx}(1-e^2)/r_B-1}{e})}$. However I do not know, if I can apply this formula to a Hohmann transfer.

b) The satellites travels $180^\circ$ in $258.8 [days]$ in the same time. The earth travels $255.2544^\circ$ and Mars travels $135.6629^\circ$. Since Mars only performs $135.6629^\circ$ we must have given him a headstart of $44.3371^\circ$. So the angle at start is $0^\circ$ between Sun and Earth and $44.3371^\circ$ between Sun and Mars. At end the angle between Sun and Earth is $255.2544^\circ$ and $180^\circ$ between Sun and Mars.
Is this correct?

c) From b) I know that Mars needs a headstart of $44.3371^\circ$. Hence I was able to calculate how many days must have been passed until Mars is in the opposition of earth.
$$\omega_E \cdot x = 44.3371 + \omega_M \cdot x \Leftrightarrow x = 94.48 [days]$$
After 94.48 days after start Mars opposes Earth. In addition I have no idea how to solve this task :).

Thank you very much for your time!

2. Nov 3, 2013

### D H

Staff Emeritus
With regard to question (a): You are overthinking the problem. The question is asking about the true anomaly of the Earth, not the satellite. You are close to having this number in your answer to (b).

With regard to question (b): The question is asking for the angle between the Sun-Earth vector and the Sun-Mars vector. It's a single number.

With regard to question (c): The question is asking when the Earth and Mars will be in opposition relative to the start of the mission. This will happen sometime between departure and arrival because Mars leads the Earth by some angle at the start but trails the Earth by some other angle at arrival.

With regard to questions (b) and (c): You are using a value for the Earth's period that is inconsistent with the distance you were given between the Sun and the Earth. Use a consistent value. Don't look it up. Calculate the Earth's period.

3. Nov 3, 2013

### Staff: Mentor

I think (a) is easier than your approach - use the known transfer time and the angular velocity of Earth.

The answer to (b) is a single number - the angle between Earth and Mars as seen from the sun. The point where the mission started is not relevant for the definition of angles.

(c) looks good.

Edit: too slow :(.

4. Nov 3, 2013

### Paul Gray

Thank you both very much for your input.

I think I understood the question b). It is asked for the angle $\angle \text{Earth }\text{Sun }\text{Mars}$. Hence it has to be as following
$$\angle ESM = \omega_E * t_h - (44.3371^\circ + \omega_M \cdot t_h) = 0.9863 \cdot 258.8 - (44.3371^\circ + 0.5242 \cdot 258.8) = 75.2749^\circ$$ Did I get it :)?

In regard to a):
We have defined the true anomaly as "the angle between perigee and the current satellite position in its orbit". The perigee defines the closest point to the object of interest and the apogee (or apocentre) the farthest point. Unfortunately I am not able to apply this knowledge to the circular orbit, we are using here.

The attached image file displays our situation. A and A' are the position of Earth and Mars at the beginning of our Hohmann transfer. B and B' display the position at the end of the transfer. From a first understanding I would guess the the true anomaly equals the answer of b) since the satellite is at Mars' position. But here I have neglected that it has to be the angle between perigee and the satellite's current position. I just took Earth position in regard to Mars.

In regard to c):
I have calculated the days necessary until Mars opposes Earth relative to the start of the Hohmann transfer (94.48 days) NOT relative to the satellites launch on Earth. I have no clue, how many days in before a satellite has to be launched to perform such a hypothetical Hohmann transfer to Mars.

Thank you again for any kind of input :)!

#### Attached Files:

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Last edited: Nov 3, 2013
5. Nov 3, 2013

### Staff: Mentor

In (a), you are asked for a change in the true anomaly. You don't have to care about an absolute value. The rate of change is just the angular speed of the object.

6. Nov 3, 2013

### Paul Gray

Thank you for your ongoing help :).

I'm not sure if it is just the angular speed of the Earth, because my exercise sheet asks for the angular speed in the exact same question, which comes after the question for the true anomaly change. So I would calculate the angular speed twice in two consecutive question.

7. Nov 3, 2013

### D H

Staff Emeritus
Close. The first two digits (75°) are correct. The rest are not. You are using inconsistent values, so your result is only good to two places. Another problem: You should only report four digits of accuracy because you were only given distances to four places.

With a circular orbit you can pick any arbitrary point as the periapsis point. Since the question is asking for the change in true anomaly, it doesn't matter what you select as that arbitrary point.

You have already determined that Mars needs to be ahead of Earth in terms of angular displacement at the start of the transfer and will be behind the Earth at the end of the transfer. Somewhere in between the Sun, Earth, and Mars will be in perfect alignment. The question is asking what time this happens, relative to the start time.