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3 years ago

6

(a) What is the escape speed on a spherical asteroid whose radius is 301 km and whose gravitational acceleration at the surface

is 0.412 m/s2

1 answer:

Allushta [10]3 years ago

7 0

Answer:

$V.E=498.02m/s^2$

Explanation:

From the question we are told that:

Radius $r=301Km$

Gravitational acceleration $g=0.412 m/s^2$

Generally the equation for Escape velocity is mathematically given by

$V.E^2=2gr$

$V.E^2=2*0.412m/s^2*301000$

$V.E^2=248024$

$V.E=\sqrt{248024}$

$V.E=498.02m/s^2$

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