What is the escape speed on a spherical asteroid whose radius is 517 km and whose gravitational acceleration at the surface is 0.636 m/s2

810.94 m/s

Explanation:

Applying,

v = √(2gR) Equation 1

Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth

From the question,

Given: g = 0.636 m/s², R = 517 km = 517000 m

Substitute these values into equation 1

v = √(2×0.636×517000)

v = √(657624)

v = 810.94 m/s

Hence, the escape velocity is 810.94 m/s

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