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Published byKeely Birkes Modified over 2 years ago

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Escape Speed Escape speed: The minimum speed required for a projectile to escape the gravitational effects of an object. For the purist: The escape speed is defined to be the speed at which all kinetic energy at the surface of the object (planet, star….) is sufficient to allow the object to escape the influence of gravity. The calculation is based on conservation of energy arguments.

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v escape = 2 G M R () 1/2 In general: Escape Speed Note: This looks like the relationship we have used in the past relating masses to orbital speeds, but watch out for the “2.”

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Show that the escape velocity for rockets leaving the surface of the earth is 11 x 10 3 m/sec. M = 5.97 x 10 24 Kg R = 6.378 x 10 6 m v escape, earth = [{2 (6.6 x 10 -11 N –m 2 /kg 2 ) (5.97 x 10 24 Kg) } / (6.378 x 10 6 m) ] 1/2 = 11 x 10 3 m/sec Escape Speed

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What is the escape velocity for a typical white dwarf? For a white dwarf, use: M = 2.18 x 10 30 Kg R = 5.1 x 10 6 m Escape Speed v escape, dwarf = [{2 (6.6 x 10 -11 N –m 2 /kg 2 ) (2.18 x 10 30 Kg) } / (5.1 x 10 6 m) ] 1/2 = 7.5 x 10 6 m/sec

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M = 3.3 x 10 30 Kg R = 2.0 x 10 4 m Escape Speed What is the escape velocity for a typical neutron star? For a neutron star, use: v escape, neutron = [{2 (6.6 x 10 -11 N –m 2 /kg 2 ) (3.3 x 10 30 Kg) } / (2.0 x 10 4 m) ] 1/2 = 1.4 x 10 8 m/sec

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Some numbers: From the surface of the earth, M = 5.97 x 10 24 Kg R = 8.756 x 10 -3 m v escape, earth = 3.0 x 10 8 m/sec Interesting piece of information: IF the earth (magic?) could be collapsed to a radius of 0.008756 m (just about a centimeter) Escape Speed v escape, dwarf = [{2 (6.6 x 10 -11 N –m 2 /kg 2 ) (5.97 x 10 24 Kg) } / (8.756 x 10 -3 m) ] 1/2

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