# Calculating the delta V budget from Earth to Mercury

## Context

While verifying a basic computation to compute the required delta V budget to get to Mercury I am experiencing some difficulties in interpreting the number 8650 m/s from Earth intercept to Mercury as given in the subway map of the solar system as presented in this question.

In addition, I noticed I do not have a crisp understanding of what is meant with "Earth intercept". Hence, I would like to present my calculation of the delta V along with the interpretation of "(Earth) Intercept" and ask if I made any mistakes in my calculations and or assumptions.

## Assumptions

• The "Earth Intercept" means the spacecraft has reached the Earth escape velocity w.r.t. Earth.
• Spherical Earth radius is approximately 6371000m.
• Earth is assumed to be a homogeneous sphere.
• LEO altitude is 250 km above the Earth surface.
• $$\mu_{Earth}=GM_{Earth}=3.98*10^14$$
• The spacecraft can reach LEO with a delta V of 9400 m/s, of which 1600 m/s are given to friction, yielding a circular velocity of 7800 m/s around leo. Verifying this with the Vis-Viva equation for $$r=a$$ (a circular orbit). $$V_{Circ-LEO}=\sqrt{\frac{\mu_{Earth}}{r_{LEO}}}=\sqrt{\frac{3.98*10^14}{6371000+250000}}=7753.18..\frac{m}{s}$$ This is considered verified, as Leo might be actually be slightly lower or higher than 250 km altitude.
• Next, the delta V required to reach Earth escape velocity from LEO is computed by first computing the escape velocity at LEO altitude using: $$V_{LEO-esc}=\sqrt{\frac{2GM}{r_{LEO}}}=\sqrt{\frac{2\mu_{Earth}}{r_{LEO}}}=\sqrt{\frac{2*3.98*10^14}{6371000+250000}}=10964.6\frac{m}{s}$$ And subtracting the previously computed circular velocity of LEO from that: $$\Delta V_{LEO-Earth-escape}=V_{LEO-esc}-V_{Circ-LEO}=10964.6-7753.18=3211.46 \frac{m}{s}$$ Which seems to be close enough with the budget to reach the Earth intercept from LEO as shown in the Solar System Subway Map. Hence, this is considered verified.
• This V-escape is interpreted as a spacecraft being able to "freely" move into any position in the Earth orbit around the Sun. In essence, it does not have enough energy to move higher or lower in its orbit around the sun, but it is "free" from Earths grip/Hill sphere. (Note this is not completely accurate I think because actually moving along Earths orbit around the Sun would require some negligible delta V).
• Since it is assumed that the "Earth Intercept" means that the spaceship is simply somewhere in the same orbit height around the sun as Earth, it is assumed that to get to Mercury, it is sufficient to reach a circular orbit equal to that of the semi-major axis (0.387 AU) of Mercuries elliptical orbit. It is fine to hit Mercury upon impact for the sake of the calculation.

## Calculations

Based on these assumptions, the actual calculation from Earth Intercept to Mercury Intercept is performed. First the Vis-Viva equations are rewritten to compute the circular velocity of Earth and Mercury:

$$$$\begin{split} V_{Earth}=\sqrt{\mu\left(\frac{2}{r_{Earth-Sun}}-\frac{1}{a_{Earth-Sun}}\right)}\\ V_{Earth}=\sqrt{\mu\left(\frac{2}{r_{Earth-Sun}}-\frac{1}{r_{Earth-Sun}}\right)}\\ V_{Earth}=\sqrt{\mu\left(\frac{1}{r_{Earth-Sun}}\right)}\\ V_{Earth}=\sqrt{\frac{\mu_{Sun}}{r_{Earth-Sun}}}\\ \end{split}$$$$ Filling in the numbers:

• $$\mu_{Sun}=1.33\cdot 10^{20} \frac{m^3}{s^2}$$
• $$r_{Earth-Sun}=1.496\cdot 10^{11} m$$ Yields: $$$$\begin{split} V_{Earth}=\sqrt{\frac{\mu_{Sun}}{r_{Earth-Sun}}}\\ V_{Earth}=\sqrt{\frac{1.33\cdot 10^{20}}{1.496\cdot 10^{11}}}=29816.73075900643\frac{m}{s}\\ \end{split}$$$$ Next, one can compute Mercury's orbital velocity around the Sun using $$r_{Mercury-Sun}=0.387\cdot 1.496\cdot 10^{11}$$:

$$$$\begin{split} V_{Mercury}=\sqrt{\frac{\mu_{Sun}}{r_{Mercury-Sun}}}\\ V_{Mercury}=\sqrt{\frac{1.33\cdot 10^{20}}{0.387\cdot1.496\cdot 10^{11}}}=47929.68129706198\frac{m}{s}\\ \end{split}$$$$

Hence the required $$\Delta V$$'s can be computed as: $$$$\Delta V_{Mercury}=V_{Mercury}-V_{Earth}=47929.68129706198-29816.73075900643=18112.95053805555\frac{m}{s}$$$$ However, this $$18112.95053805555\frac{m}{s}$$ is not in the neighbourhood of the $$8650\frac{m}{s}$$ shown for the Earth intercept to Mercury Intercept.

## Code of calculations

For completeness, here is the Python code that performed the actual computations from Earth intercept to Mercury Intercept:

import math

# Initialize parameters:
mu_sun=1.33*10**20
r_earth_sun=1.496*10**11
r_mercury_sun_au=0.387
r_mercury_sun=r_mercury_sun_au*r_earth_sun

# Compute orbital velocities
v_earth=(mu_sun/r_earth_sun)**0.5
print(f'v_earth=\n{v_earth}')

v_mercury=(mu_sun/r_mercury_sun)**0.5
print(f'v_mercury=\n{v_mercury}')

dv_mercury=v_mercury-v_earth
print(f'dv_mercury=\n{dv_mercury}')


## Question

What did I do wrong in my computation of the Earth Intercept to Mercury Intercept?

I see two main things throwing off your calculations:

1. You can not simply subtract the velocity of one planet from another to get interplanetary transfer costs.

An optimal transfer consists of an elliptic orbit touching the orbit of the inner planet at perihelion, and the outer planet at aphelion.

Thus, the numbers you should try to obtain are

• the difference in velocity between Earth and the aphelion velocity of such an ellipse.
• the difference in velocity between Mercury and the perihelion velocity of such an orbit.

Since you already have command of the Vis-Viva equation, you should be able to compute both of them.

1. Waiting until after escape from Earth (and Mercury) to perform the interplanetary transfer is inefficient.

That is, you don't simply add up the two numbers from the previous step. Instead, both the escape impulse and the transfers are made together at LEO (and LMO) altitude.

The key equation being:

$$v^2 = v_e^2 + v_{\infty}^2$$

Where $$v$$ is the velocity at LEO after the impulse is performed, $$v_e$$ is the escape velocity and $$v_{\infty}$$ is the velocity relative to Earth after escape.

Worked example of the Earth side of things. You would have to do the same thing on Mercury to reduce the flyby to an intercept.

$$v_{aphelion} = \sqrt{\mu_{Sun}\left(\frac{2}{a_{Earth}} - \frac{2}{a_{Earth} - a_{Mercury}}\right)} \approx 22250 m/s$$

This is a relative velocity to Earth of $$7533 m/s$$

We can then use:

$$v^2 = v_e^2 + v_{\infty}^2$$

$$v^2 = (10964.6m/s)^2 + v_{\infty}^2$$

Insert the desired relative velocity after escape:

$$v^2 = (10964.6 m/s)^2 + (7533 m/s)^2$$

And then solve:

$$v = 13303 m/s$$

By subtracting the circular velocity, we have $$\Delta v_1 = 5550 m/s$$

You now to the same thing for arriving at Mercury for the other half of the cost.

• Is the following an accurate understanding of the overall implication of your feedback: I should compute the Delta V using a Hohmann transfer to Mercury, where the velocity at Aphelion is that of Mercuries circular velocity around the Sun, instead of computing the difference in circular orbits? – a.t. Apr 19 at 12:10
• @a.t.Yes, you would have to use a proper Hohmann transfer, although I have tried to be careful in saying what parts of the transfer should be included. (and "velocity at Aphelion is that of Mercuries circular velocity around the Sun" does not make sense). – SE - stop firing the good guys Apr 19 at 12:14