<p>Olympus Mons is fairly close to the equator, so the radius at its latitude is only a little under the equatorial radius.</p>
<p>According to Wikipedia <a href="https://en.wikipedia.org/wiki/Olympus_Mons" rel="nofollow noreferrer">Olympus Mons</a>, its altitude is 21.9 km, and its geodetic latitude is 18.65° N. Hence its <a href="https://en.m.wikipedia.org/wiki/Latitude#Parametric_latitude_(or_reduced_latitude)" rel="nofollow noreferrer">parametric latitude</a> is ~18.5477°, and the radius of the ellipsoid at that latitude is ~3394.18 km. So at the top of Olympus Mons, the radius is ~3416.1 km.</p>
<hr />
<p>Here's some of the relevant algebra. The geodetic latitude is the usual latitude, which is defined in terms of the surface normal. But it's often easier to do calculations with the parametric latitude.</p>
<p>For an ellipse with semi-major axis <span class="math-container">$a$</span> and semi-minor axis <span class="math-container">$b$</span>, the parametric latitude <span class="math-container">$\beta$</span> and the geodetic latitude <span class="math-container">$\phi$</span> are related by:
<span class="math-container">$$\tan\beta = \frac{b}{a}\tan\phi$$</span>
The coordinates of a point on the ellipse are given by the parametric equations:
<span class="math-container">$$x = a\cos\beta$$</span>
<span class="math-container">$$y = b\sin\beta$$</span>
So the radius is given by
<span class="math-container">$$r^2=a^2\cos^2\beta + b^2\sin^2\beta$$</span></p>
<p>The corresponding equation in terms of the geodetic latitude is:
<span class="math-container">$$\frac{1}{r^2} = \frac{\sin^2\phi}{a^2} + \frac{\cos^2\phi}{b^2}$$</span></p>

CC BY-SA 4.0
2024-09-06T23:04:13.940Z
2024-09-07T03:10:29.763Z
2024-09-07T03:10:29.763Z
PM 2Ring
<p>According to <a href="https://www.google.com/search?q=mars%20mass%20compared%20to%20earth&amp;sca_esv=24731a508b288d28&amp;source=hp&amp;ei=bEPbZrqeCKXO1sQP1aGX6AY&amp;iflsig=AL9hbdgAAAAAZttRfKSuAKqmYDxWSO3-ZAYqsb2wWL1k&amp;ved=0ahUKEwi6hKOM866IAxUlp5UCHdXQBW0Q4dUDCA4&amp;uact=5&amp;oq=mars%20mass%20compared%20to%20earth&amp;gs_lp=Egdnd3Mtd2l6IhttYXJzIG1hc3MgY29tcGFyZWQgdG8gZWFydGgyBxAAGIAEGBMyCBAAGBMYFhgeMggQABgTGBYYHjIIEAAYExgWGB4yCBAAGBMYFhgeMggQABgTGBYYHkiBIlAAWOAgcAB4AJABAZgBtQWgAYkoqgEMMy4xNS40LjQuMC4xuAEDyAEA-AEBmAIaoAK9JsICChAAGIAEGEMYigXCAgsQABiABBixAxiDAcICCxAuGIAEGLEDGIMBwgIOEC4YgAQYsQMYgwEYigXCAgUQLhiABMICCBAuGIAEGLEDwgIOEAAYgAQYsQMYgwEYigXCAhYQLhiABBixAxjRAxhDGIMBGMcBGIoFwgINEC4YgAQYsQMYQxiKBcICEBAAGIAEGLEDGEMYgwEYigXCAgUQABiABMICCxAuGIAEGMcBGK8BwgIQEC4YgAQYsQMYQxiDARiKBcICCBAAGIAEGLEDwgILEC4YgAQY0QMYxwHCAhEQLhiABBixAxjRAxiDARjHAcICCBAAGBYYChgewgIGEAAYFhgewgIIEAAYFhgeGA_CAgoQABgTGBYYHhgPwgIIEAAYgAQYogSYAwCSBwwwLjE4LjQuMy4wLjGgB5TYAQ&amp;sclient=gws-wiz" rel="nofollow noreferrer">Mars mass compared to earth</a>,</p>
<p>the Equatorial radius of Mars is 3396.2 kms and the Polar radius is 3376.2 . But I want to know the radius of Mars over Mount Olympus and it wont be the same as any of those two.</p>
<p>What's the radius of Mars over Mount Olympus?</p>

CC BY-SA 4.0
2024-09-07T08:39:52.450Z
I'm curious why you want to know this radius. Is it in connection to your question https://space.stackexchange.com/q/66839/38535 ? If so, then the radius isn't really relevant, you just need to know the altitude above the ellipsoid (the Mars equivalent of sea level). See https://en.wikipedia.org/wiki/Geopotential Also see this question on the gravity of a rotating ellipsoid: https://physics.stackexchange.com/q/8074/123208
PM 2Ring
CC BY-SA 4.0
2024-09-07T12:47:03.470Z
@PM2Ring I wanted to know if Mount Olympus is the best place to have a launching facility on Mars, considering escape velocity, since escape velocity is lower when the radius is larger. But according to the answer, it's the equator
Pablo
CC BY-SA 4.0
2024-09-07T22:11:34.880Z
@Pablo I suspect that any gains in delta-v gained by the extra handful of km will be dwarfed by the additional difficulties regarding putting a launch facility on top of a 20 km tall mountain.
fyrepenguin
CC BY-SA 4.0
2024-09-07T23:14:34.717Z
@fyrepenguin I don't think it's a steep mountain like those on Earth
Pablo
<mars><radius><olympus-mons>
What's the radius of Mars over Mount Olympus?
Pablo
# What's the radius of Mars over Mount Olympus?

(Asked by: username_0 on 2024-09-06)

According to [Mars mass compared to earth](https://www.google.com/search?q=mars%20mass%20compared%20to%20earth&sca_esv=24731a508b288d28&source=hp&ei=bEPbZrqeCKXO1sQP1aGX6AY&iflsig=AL9hbdgAAAAAZttRfKSuAKqmYDxWSO3-ZAYqsb2wWL1k&ved=0ahUKEwi6hKOM866IAxUlp5UCHdXQBW0Q4dUDCA4&uact=5&oq=mars%20mass%20compared%20to%20earth&gs_lp=Egdnd3Mtd2l6IhttYXJzIG1hc3MgY29tcGFyZWQgdG8gZWFydGgyBxAAGIAEGBMyCBAAGBMYFhgeMggQABgTGBYYHjIIEAAYExgWGB4yCBAAGBMYFhgeMggQABgTGBYYHkiBIlAAWOAgcAB4AJABAZgBtQWgAYkoqgEMMy4xNS40LjQuMC4xuAEDyAEA-AEBmAIaoAK9JsICChAAGIAEGEMYigXCAgsQABiABBixAxiDAcICCxAuGIAEGLEDGIMBwgIOEC4YgAQYsQMYgwEYigXCAgUQLhiABMICCBAuGIAEGLEDwgIOEAAYgAQYsQMYgwEYigXCAhYQLhiABBixAxjRAxhDGIMBGMcBGIoFwgINEC4YgAQYsQMYQxiKBcICEBAAGIAEGLEDGEMYgwEYigXCAgUQABiABMICCxAuGIAEGMcBGK8BwgIQEC4YgAQYsQMYQxiDARiKBcICCBAAGIAEGLEDwgILEC4YgAQY0QMYxwHCAhEQLhiABBixAxjRAxiDARjHAcICCBAAGBYYChgewgIGEAAYFhgewgIIEAAYFhgeGA_CAgoQABgTGBYYHhgPwgIIEAAYgAQYogSYAwCSBwwwLjE4LjQuMy4wLjGgB5TYAQ&sclient=gws-wiz),

the Equatorial radius of Mars is 3396.2 kms and the Polar radius is 3376.2 . But I want to know the radius of Mars over Mount Olympus and it wont be the same as any of those two.

What's the radius of Mars over Mount Olympus?

**Tags:** mars, radius, olympus-mons

> **Comments:**
> **username_1**: I'm curious why you want to know this radius. Is it in connection to your question https://space.stackexchange.com/q/66839/38535 ? If so, then the radius isn't really relevant, you just need to know the altitude above the ellipsoid (the Mars equivalent of sea level). See https://en.wikipedia.org/wiki/Geopotential Also see this question on the gravity of a rotating ellipsoid: https://physics.stackexchange.com/q/8074/123208  
> **username_0**: @username_1 I wanted to know if Mount Olympus is the best place to have a launching facility on Mars, considering escape velocity, since escape velocity is lower when the radius is larger. But according to the answer, it's the equator  
> **username_2**: @username_0 I suspect that any gains in delta-v gained by the extra handful of km will be dwarfed by the additional difficulties regarding putting a launch facility on top of a 20 km tall mountain.  
> **username_0**: @username_2 I don't think it's a steep mountain like those on Earth  

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## 1 answers

### Answer by username_1 on 2024-09-06. Score: 8 | Accepted answer

Olympus Mons is fairly close to the equator, so the radius at its latitude is only a little under the equatorial radius.

According to Wikipedia [Olympus Mons](https://en.wikipedia.org/wiki/Olympus_Mons), its altitude is 21.9 km, and its geodetic latitude is 18.65° N. Hence its [parametric latitude](https://en.m.wikipedia.org/wiki/Latitude#Parametric_latitude_(or_reduced_latitude)) is ~18.5477°, and the radius of the ellipsoid at that latitude is ~3394.18 km. So at the top of Olympus Mons, the radius is ~3416.1 km.

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Here's some of the relevant algebra. The geodetic latitude is the usual latitude, which is defined in terms of the surface normal. But it's often easier to do calculations with the parametric latitude.

For an ellipse with semi-major axis $a$ and semi-minor axis $b$, the parametric latitude $\beta$ and the geodetic latitude $\phi$ are related by:
$$\tan\beta = \frac{b}{a}\tan\phi$$
The coordinates of a point on the ellipse are given by the parametric equations:
$$x = a\cos\beta$$
$$y = b\sin\beta$$
So the radius is given by
$$r^2=a^2\cos^2\beta + b^2\sin^2\beta$$

The corresponding equation in terms of the geodetic latitude is:
$$\frac{1}{r^2} = \frac{\sin^2\phi}{a^2} + \frac{\cos^2\phi}{b^2}$$


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