Posted by Davin Flateau on 9 Aug 2005 at 10:38 am.
Filed under Astronomy.
I’ve had more than a few people come up to me recently asking about an email forwarded to them by a friend. The email got their attention because it talks about an upcoming spectacular view of Mars. Maybe you’ve seen the email, which includes the claims:
“The Red Planet is about to be spectacular.”
“Earth is catching up with Mars [for] the closest approach between the two planets in recorded history.”
“On August 27th … Mars will look as large as the full moon.”
“NO ONE ALIVE TODAY WILL EVER SEE THIS AGAIN.”
Many people have told me they’ve heard something about this on the news, which makes the situation even more confusing. Well, here at Stars Over Kansas, we’re here to set the record straight!
It’s a hoax.
Now this Halloween, Mars IS going to get a bit closer than it usually does - 43 million kilometers (69 million kilometers) from the Earth. This will make Mars look like a very bright star in the night sky, and make a nice red-orange pinprick of light for trick-or-treaters to look up at. Certainly not as big as the full moon!
It looks like the hoaxsters twisted some facts from 2003, when Mars got as close as its going to get to Earth in 60,000 years or so. Back then it came to within 34 million miles (56 million kilometers ) - almost as close as it’s going to get on October 30. Here at Exploration Place, thousands lined up to get an unprecedented look at Mars. It will be just about as good this time, so go out to your local observatory and get a good look.
It’s going to be a great show of Mars, but I hope people won’t be too disappointed that it won’t be close enough to wreak havoc on the Earth.
By the way - I’ve decided to offer a special prize to the first person from Wichita who can calculate how far Mars would have to be from the Earth to appear as big as the full moon (0.5 degrees). Post your best calculations in the comments!
Davin Flateau
Jim on 9 Aug 2005 at 11:45 am: 1
Where did I put my trigonometry textbook.
Jonathan D. Wells on 12 Jan 2006 at 12:57 am: 2
I came across your site and this article while looking for MWG maps. I’m not from Wichita, Kansas or the U.S. but I can’t resist a good problem.
Can’t draw anything here so I’ll try to explain verbally.
Draw a circle (Mars) and then draw a right triangle with one vertex in the center of the circle, another at the top of the circle and the last one out to the left of the circle (viewer on Earth’s surface) forming a horizontal line to the center.
The small vertical side represents the radius of Mars, 3400 km. The angle at the center vertex is 90 degrees (which incedently simplifies the trigonometry) and the small angle at the viewer is 0.25 degrees because we’ve set the diameter of Mars to 0.5 degrees to match the Moon.
The horizontal line is the distance between Mars and the viewer on Earth, which is what we’re trying to find. Since this is a right triangle we use the equation:
TAN(x) = Opposite Side (O) / Adjacent Side (A)
A = O / TAN(x)
A = 3400 km / TAN(0.25)
A = 3400 km / 0.004363
A = 779,280 km
So Mars would appear to be the same size as the Moon if it was approximately 780,000 km from the Earth.
Interestingly enough, this distance is practically EXACTLY 100 times closer than the average distance between Mars and Earth, 78,000,000 km.