I am just wondering at what speed does Tesla Roadster is currently travelling toward the Asteroid Belt (Relative to Earth)?
3 Answers
I've now created a website that will populate the speed in near real time, http://www.whereisroadster.com/. As of writing this, it states:
The current location is 332874 miles (535709 km) from Earth, moving away from Earth at a speed of 8243 miles/hour (13266 km/hour).
Also note that it isn't actually going to enter the Asteroid belt, that was an issue with Elon's initial tweet. No idea why he was wrong, but he was wrong.
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2$\begingroup$ It would be good to include the information that this velocity is with respect to the Earth on the website. With respect to the Sun, or the Solar System barycenter, it's moving roughly 10x faster. This reminds me to fix my answer, since the OP does ask with respect to the Earth! (blush) $\endgroup$– uhohFeb 8, 2018 at 15:01
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1$\begingroup$ @uhoh Good call, I've added a bit of clarification on that point. $\endgroup$– PearsonArtPhoto ♦Feb 8, 2018 at 15:18
EDIT (new):
Musk's tweet now appears to be wrong according to this comment! Now recalculating.
Using JPL's Horizons data for the planets and Roadster as I discuss more fully in this answer, I get the following:
min max
Roadster: 19.9 33.9 km/sec
Earth: 29.3 30.3 km/sec
Mars: 22.0 26.6 km/sec
You can see that the Roadster's velocity with respect to the Solar System barycenter has a wide swing, in accordance with it's more elliptical orbit.
With respect to the Earth, things are a lot different. Right now it's only about 3.5 km/s, and will drop down to about 3.2 km/s in a few weeks. There's a nice plot in this answer.
The following is obsolete!
Elon Musk's tweet shows the figure below, indicating a semi-major axis of about (2.61+0.98)/2 = 1.795 AU. Mars' semi-major axis is about 1.524 AU.
One AU is 149597870700. meters, and the standard gravitational parameter of the Sun $GM$ is about 1.327E+20 m^3/s^2. Using the values for perihelion and aphelion shown in the plot, and the vis-viva equation:
$$v^2 = GM \left( \frac{2}{r}-\frac{1}{a}\right),$$
I get a velocity at perihelion (which is at least darn close to where it is now) of 36,300 m/s, and at aphelion of about 13,600 m/s, which will happen in about 880 days, based on the two decimal place accuracy of the numbers shown in the tweet.
edit: Scott Manley gets a period of 910 days (2.404 years) at 05:40
in his new video (linked in this answer) rather than 880 days. I'm not sure of the reason for the discrepancy though. Perhaps he used the $C_3$ instead? In any even, he certainly knows a lot about asteroids!
Bothered by the periapsis being only 0.98 AU? I was. Then I remembered that the Earth is closest to the Sun during the Northern Hemisphere's winter.
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1$\begingroup$ It's after midnight. Please someone check my arithmetic, correct if necessary. $\endgroup$– uhohFeb 7, 2018 at 16:30
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1$\begingroup$ But as there is one more ∆v left. So don't you think it will increase the speed of the Roadster? Or you are also considering that in your solution. $\endgroup$ Feb 7, 2018 at 16:38
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1$\begingroup$ @SudhanshuGaur The green line shows the projected orbit for the next two years, and it comes from SpaceX. I'm sure its after the 3rd and final burn, which happened many hours ago already. The tweet itself actually says "Third burn successful. Exceeded Mars orbit and kept going to the Asteroid Belt." $\endgroup$– uhohFeb 7, 2018 at 16:47
I plotted the HORIZONS data (SpaceX Roadster (spacecraft) (Tesla) [-143205]
) over time since launch in order to answer the question for the near future:
The red line gives the answer to your question, by date (top axis labels) or days since launch (bottom axis labels). The velocity of Earth is taken from [399]
(Earth [Geocenter]).
For reference, the blue line gives the velocity of the spacecraft relative to the solar system barycenter (SSB), and the green dashed line shows the velocity of Earth relative to the SSB.
25,300
?? $\endgroup$