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?
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.
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.
- 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.
- 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.
- @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.
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: plt.figure() 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.title(title) plt.xlabel('time (min)') plt.show()