The escape velocity for any body may be calculated from Ve = sqrt[2µ/r] where Ve = the escape velocity in feet per second, µ = the gravitational constant of the body (1.407974x10^16 ft.^3/sec.^2 for Earth) and r = the surface distance in feet. For earth, with an equatorial radius of ~3963 miles r becomes 3963(5280) = ~20,924,640 feet and Ve = sqrt[2(1.407974x10^16)/20,924,640] = ? ft./sec. Divide your answer by 1.467 to get the answer in miles per hour.
For the moon, µ = 1.731837x10^14 ft.^3/sec.^2, r = 1080 miles = 5,702, 400 feet.
For the sun, µ = 4.68772x10^21 and r = 432,495 miles = 2,283,573,600 feet.
How fast does a rocket need to be going to break free of Earth's gravity? How fast would it have to be going in order to completely leave the solar system? (For the second part of this problem, assume that the Sun is the only mass in the solar system and that the rocket is beginning its journey from Earth's orbit?)
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