Virgin Galactic’s WhiteKnight2 and SpaceShipTwo soar above the runway at the New Mexico Spaceport. Credit: Mark Greenberg
divided by the hypotenuse, and sine is defined as the
opposite side divided by the hypotenuse, we can write
the two trigonometric equations using this information.
cosØ = DIST / GD
sinØ = ALT / GD
Solving for Distance and Altitude, we get
DIST = GD * cosØ
ALT = GD * sinØ
We can now calculate the altitude and the
distance to the Spaceport for our spacecraft.
We must first convert 15 miles to meters,
which comes to about 24,140 m. Therefore,
ALT = 24140 * sin(35) = 13,846 m
DIST = 24140 * cos(35) = 19,774 m
Conclusion
When the spacecraft is about 15 miles distant, it is
at an altitude of 13,846 m (8.6 mi) above the ground,
and is 19,774 m (12.3 mi) from Spaceport America.
Is there a way that we can make a quick check
of our results? Sure there is! Use the Pythagorean
Theorem. Leaving the units in miles, we get:
Note: we will be using degrees instead of
radians to keep things a little simpler. We will, of
course, always continue to convert to S.I. units.
a2 + b2 = c2
(8.6)2 + (12.3)2 ?= (15)2 = 225
(73.96) + (151.29) = 225.25
Example
The result is close enough to 15 squared. We
therefore are on track to a safe landing!
The Landing Laser at this moment in time reads a
Glide Slope of 35 degrees with the spacecraft at 15
miles distant. What is the altitude and ground distance
from the Spaceport of the returning spacecraft?
S.T.E.M. education...
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