How long does it take for a signal to travel between Earth, and Juno at Jupiter?

Question

How long would it take for Juno's communication to reach the Earth? In other words: what is the time delay between perception by Juno and perception by NASA researchers?

Answer 1

Using NASA's Eyes measuring the distance from Jupiter to Earth at this moment (5th Jul 2016, 11:50 CEST) is 48 light minutes, 21.39 light seconds, and that would be the time Juno's communications take to reach Earth.

Answer 2

EDIT: based on @Beska's comment, I went back and calculated the difference including light time. In other words, you have to use Jupiter's position roughly 48 minutes ago to state the travel time. Using the observe() method, which does this, there is a difference of 0.02 seconds. This doesn't really matter, considering that Juno is in a large orbit around Jupiter, not inside Jupiter - yet! :)

import numpy as np
import matplotlib.pyplot as plt

from skyfield.api import load

data   = load('de421.bsp')
ts     = load.timescale()
t      = ts.utc(2016, 7, 5, 9, 50, 0)

jupiter, earth  = data['Jupiter barycenter'], data['Earth']
jpos, epos      = jupiter.at(t).position.km, earth.at(t).position.km
d_instantaneous = np.sqrt(((jpos - epos)**2).sum())

d_light = earth.at(t).observe(jupiter).distance().km  # where WAS Jupiter 48 minutes ago?

clight = 299792.458  # km/s

print "d_instantaneous / c = ", d_instantaneous/clight
print "d_light / c =         ", d_light/clight

gives

d_instantaneous / c =  2901.39437989  
d_light / c =          2901.4127772

So it looks like NASA Eyes is using the simpler method of using instantaneous positions, and not actually back-calculating the position where Jupiter WAS when the signal would have started.

---

This is a different way to look at it. This is the variation of distance, light time, and also angular separations from the sun for Earth seen from Jupiter, and Jupiter seen from Earth. When they are too close, radio communication can become difficult.

I used Python and the Skyfield package. @SF. gives the right answer, I'm just plotting values as a function of time. The way I chose to do this I didn't used Skyfield's methods for light-time corrections, gravitation or astronomical aberration or atmospheric refraction, which are done with the .observe() and .apparent() methods. (Not all of those affect light time significantly anyway.) For this kind of rough plot it's not necessary so I used a short-cut.

The black dot is approximately July 4, 2016 for reference.

import numpy as np
import matplotlib.pyplot as plt

from skyfield.api import load

data = load('de421.bsp')

years = np.linspace(2015, 2020, 1000)

ts     = load.timescale()
t      = ts.utc(years, 0, 0)

jupiter = data['Jupiter barycenter']
earth   = data['Earth']
sun     = data['sun']

jpos = jupiter.at(t).position.km
epos = earth.at(t).position.km
spos = sun.at(t).position.km

d_je = np.sqrt(((jpos-epos)**2).sum(axis=0))
d_js = np.sqrt(((jpos-spos)**2).sum(axis=0))
d_es = np.sqrt(((epos-spos)**2).sum(axis=0))

clight = 2.9979E+05  # km/sec speed of light

t_je, t_js, t_es = [thing/clight for thing in [d_je, d_js, d_es]]

# dot products for angles
sep_js = np.arccos( ((jpos-epos)*(spos-epos)).sum(axis=0) / (d_je*d_es))
sep_es = np.arccos( ((epos-jpos)*(spos-jpos)).sum(axis=0) / (d_je*d_js))

degs = 180. / np.pi
ttjly4 = ts.utc(2016, 7, 4).tt
i = np.argmax(t.tt>ttjly4)  # find the index of the first time point after 4 July 2016

fig = plt.figure()

ax = fig.add_subplot(2, 2, 1)
ax.plot(years, d_je)
ax.plot(years[i], d_je[i], 'ok')
ax.set_title("Jupiter-Earth distance(km)")
ax.ticklabel_format(useOffset=False)

ax = fig.add_subplot(2, 2, 2)
ax.plot(years, t_je/60.)
ax.plot(years[i], t_je[i]/60., 'ok')
ax.set_title("Jupiter-Earth light-time (minutes)")
ax.ticklabel_format(useOffset=False)

ax = fig.add_subplot(2, 2, 3)
ax.plot(years, degs*sep_js )
ax.plot(years[i], degs*sep_js[i], 'ok' )
ax.set_title("Jupiter-Sun separation @Earth (deg)")
ax.ticklabel_format(useOffset=False)

ax = fig.add_subplot(2, 2, 4)
ax.plot(years, degs*sep_es )
ax.plot(years[i], degs*sep_es[i], 'ok' )
ax.set_title("Earth-Sun separation @Jupiter (deg)")
ax.ticklabel_format(useOffset=False)

plt.show()

Answer 3

Jupiter is about 5.2 A.U. from the sun and earth is 1 A.U. from the sun. So distance between Jupiter and Earth ranges from 4.2 to 6.2 A.U.

1 A.U. takes about 500 seconds for light to traverse. So light's travel time from earth to Jupiter takes 2100 to 3100 seconds which is 35 to 52 minutes.

Answer 4

The space.com folks assert the closest earth and jupiter can get in their elliptical paths is 588 million kilometers. My physics text states the speed of light is 1002 million kilometers per hour. Division yields decimal hours, then multiplication to get minutes: approximately 35 minutes. Since the actual distance between Jupiter and the Earth the day after Juno arrived is probably a bit more than the minimum I started with, so radio connect is a minimum 35 minutes.

Answer 5

Minimum distance of Jupiter from Earth (space.com) $\approx588*10^6 \text{km}$

Maximum distance of Jupiter from Earth (space.com) $\approx968*10^6 \text{km}$

$c \approx 3.0*10^8 m/s$

Minimum time for signal to reach Earth:

$$588*10^6 \text{km} * (1000m / 1\text{km}) * (1/c) * (1\text{min} / 60\text{sec}) \approx 32.67 \text{min}$$

Maximum time for signal to reach Earth:

$$968*10^6 \text{km} * (1000\text{m} / 1\text{km}) * (1/c) * (1\text{min} / 60\text{sec}) \approx 53.78 \text{min}$$