calculate the acceleration of gravity on the surface of kepler-62e Kepler-62e is an exoplanet that orbits within the habitable z
one around its parent star. The planet has a mass that is 3.57 times larger than Earth's and a radius that is 1.61 times larger than Earth's.
1 answer:
The acceleration of gravity is proportional to the planet's mass, and inversely proportional to the square of the planet's radius.
So on this particular planet, the acceleration of gravity will be
(Earth gravity) x ( 3.57 / 1.61²)
That's (9.81 m/s²) x (1.377) = <em>13.51 m/s²</em>
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Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
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i think it's 1400
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w = Force x displacement
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