Earth travels around the Sun at 67,000 mph. Hypothetically if it was possible to be in Space in a ship or Space walk and be at a complete stop and stationary outside of Earth’s orbit and not affected by its gravity, wouldn’t Earth look like a blur as it goes by?

  • $\begingroup$ Astronauts on ISS see Earth moving below them at velocity about 7.5 km/s from height of 400 km. So if we will make a video from ISS 4 times faster - we'll see how it would look. It will be fast, but not blurring, even from relatively close distance of 400 km. $\endgroup$ – Heopps Sep 28 '20 at 7:40
  • $\begingroup$ A reminder that I think should become obligatory when questions refer to stopping: "a complete stop" always depends on your point of view. You can only match speeds with something else (in this case the sun). $\endgroup$ – Erin Anne Sep 30 '20 at 2:03

Firstly, it's going to take a big rocket to do this (larger than any rocket we have built so far btw). We're travelling together with the Earth, so our speed relative to the Earth is 0. Thus, this isn't really "coming to a halt", it's more like "speeding up" to 67,000 mph (30 km/s).

Does the Earth blur past at 30 km/s? Not really.

Earth is 12,756 km across, so it would take 7 minutes for the Earth to move 1 diameter over. If you had an equally sized bowling lane for the Earth to roll on as a bowling ball, it would need 12 hours to reach the pins.

  • $\begingroup$ +1 I was surprised to see this result at first, but it's undeniable. While answers to 1, 2 show to what lengths it is necessary to deblur images of Earth's surface taken from LEO, those are high resolution images taken through long focal length optics. $\endgroup$ – uhoh Sep 27 '20 at 23:37
  • $\begingroup$ I like the way you put that, it's easy to visualize. $\endgroup$ – GdD Sep 29 '20 at 7:33

Make the position of the Earth at the midpoint in time (t=0) to be the origin.

The sun is now at [-1 AU, 0].

Move the Earth backwards from the origin by two Earth diameters (25,512 km).

Deposit a floating astronaut at [0, +2 Earth diameters] with zero velocity , but deny them any means of propulsion per the question's premise.

Now start and see what happens as the earth moves from two diameters before to two diameters after.


  1. Astronaut needs no propulsion, for this half-hour ordeal, acceleration due to the Sun's and Earth's gravity only manages to move them by about 640 kilometers. They will of course fall into the Sun 63 days later.
  2. At 25,000 km the Earth/s 29 km/s looks like less than 0.1 degrees per second. While answers to the following show to what lengths it is necessary to deblur images of Earth's surface taken from only a few hundred km above the Earth's surface, even those are high resolution images taken through long focal length optics.
  3. @SE - stop firing the good guys's answer is correct! Also, I've assumed that aliens have placed us at zero velocity rather than Earth's biggest rocket ever.

we are drifting while Earth passes us by

Python script:

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

def deriv(X, t):
    xe, xp, ve, vp = X.reshape(4, -1)
    aes = -GMs * (xe-xs) * ((xe-xs)**2).sum()**-1.5
    aps = -GMs * (xp-xs) * ((xp-xs)**2).sum()**-1.5
    ape = -GMe * (xp-xe) * ((xp-xe)**2).sum()**-1.5
    return np.hstack([ve, vp, aes, aps+ape])

GMe, GMs = 3.986E+14, 1.327E+20

a = 150E+06 * 1000 # ~1 AU
d = 2 * 2 * 6378137

xs = np.array([-a, 0])

vorbit = np.sqrt(GMs/a)

xe = np.array([np.sqrt(a**2 - d**2) - a, -d])
ve = vorbit * np.array([d/a, np.sqrt(1**2 - (d/a)**2)])

xp, vp = np.array([d, 0]), np.array([0, 0])

X0 = np.hstack([xe, xp, ve, vp])
T = 2 * d / vorbit  # time for Earth to "swoosh" from -d to +d
times = np.arange(-T/2, T/2, 60) # plot at 1 minute intervals

answers, info = ODEint(deriv, X0, times, full_output=True)

if True:
    titles = 'Earth (km)', 'Person (km)', 'Earth (km/s)', 'Person (km/s)'
    things = list(zip(answers.T.reshape(4, 2, -1), titles))
    for i, ((x, y), title) in enumerate(things[:2]):
        plt.subplot(1, 2, i+1)
        plt.plot(times/60., x/1000.)
        plt.plot(times/60., y/1000.)
        plt.ylim(-1.02*d/1000., 1.02*d/1000.)
        plt.xlabel('time (min)')

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