Japanese Probe Returns First HD Video of the Moon 206
Riding with Robots writes "The Kaguya probe, now in lunar orbit, has sent down the first footage of the moon's surface from its onboard high-definition TV camera. The Kaguya mission, which consists of a main orbiter and two smaller satellites in a 100-km-high, polar orbit, is slated to officially begin its science phase in December."
Re:I don't see any stars is this a fake video of t (Score:5, Informative)
Just in case of "idiot": Stars are not that bright and require a longer exposure time. The moon is actually very bright and a short exposure time is required to prevent the image from being washed out. The result is the stars don't appear because the camera has been desensitised to take clear pictures of bright things.
Take a picture of any brightly lit object on a starry night and you'll likely find the stars aren't in the photo.
=Smidge=
P.S. Whoever gave the parent an "Insightful" mod was definately an idiot.
Re:I don't see any stars is this a fake video of t (Score:4, Informative)
Where's the HD? (Score:4, Informative)
The press release (Score:5, Informative)
The Japan Aerospace Exploration Agency (JAXA) and NHK (Japan Broadcasting Corporation) have successfully performed the world's first high-definition image taking by the lunar explorer "KAGUYA" (SELENE,) which was injected into a lunar orbit at an altitude of about 100 km on October 18, 2007, (Japan Standard Time. Following times and dates are all JST.)
The image shooting was carried out by the onboard high definition television (HDTV) of the KAGUYA, and it is the world's first high definition image data acquisition of the Moon from an altitude about 100 kilometers away from the Moon.
The image taking was performed twice on October 31. Both were eight-fold speed intermittent shooting (eight minutes is converged to one minute.) The first shooting covered from the northern area of the "Oceanus Procellarum" toward the center of the North Pole, then the second one was from the south to the north on the western side of the "Oceanus Procellarum." The moving image data acquired by the KAGUYA was received at the JAXA Usuda Deep Space Center, and processed by NHK.
The satellite was confirmed to be in good health through telemetry data received at the Usuda station.
mod parent up (Score:5, Informative)
Re:Real Time? (Score:3, Informative)
Re:Note to JAXA (Score:5, Informative)
Titled: "Ocean of Storms", western side
Filming Start Time: 20:51 (GMT) October 30, 2007
Filming End Time: 20:59 (GMT) October 30, 2007
Filming Start Location: 250degN, 275-282deg E
Filming End Location: 49deg N, 275-283deg E
"Kaguya" altitude: 110km (68 mi) above moon surface
8 minutes of footage edited down to 1 minute (8x speed).
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Located in the left section of the Northern hemisphere of the moon, the western-most part of "Ocean of Storms" was filmed while navigating a South to North direction. The right side of the screen show the shadowy "dark oceans" and the sun-lit areas are the highlands.
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"Kaguya" HD Camera footage.
Titled: Lunar North Pole
Filming Start Time: 19:07 (GMT) October 30, 2007
Filming End Time: 19:15 (GMT) October 30, 2007
Filming Start Location: 66deg N, 274-288deg E
Filming End Location: 87deg N, 26-161deg E
"Kaguya" altitude: 110km (68 mi) above moon surface
8 minutes of footage edited down to 1 minute (8x speed).
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The following was filmed while navigating from the northern area of "Ocean of Storms" to the north pole of the Moon. Due to the higher latitude of this voyage, the lower angle of the sun caused the crater formations to cast long shadows.
Youtube link (Score:5, Informative)
Re:Speed? (Score:1, Informative)
E = v^2/2 - mu/r = -mu/(2*a)
E is the energy, mu is the gravitational constant, G, multiplied by the mass of the moon. a is the semi-major axis of the orbit, which is just the radius of the orbit if we assume it's circular (which may or may not be the case). Then
v_circ^2/2 - mu/r = -mu/(2*r)
Which becomes
v_circ = sqrt(mu/r)
For the moon, mu=4.903e12 m^3/s^2. The radius, r, would be the moon's radius, 1738km, plus the altitude of the orbit, 50km for example.
v_circ = 1.656 km/s
Then for a circular orbit the orbital period, T, is
T = (2*pi*r) / v_circ = 6784 s = 113 minutes.
I know no one cares, but I didn't TA astrodynamics for nothing. I live for this crap.