Question

The escape velocity of earth is 11.2 km/s. Find the orbital velocity of the smallest possible orbit

Answer 1

Answer:

The orbital speed can be found using v = SQRT(G*M/R). The R value (radius of orbit) is the earth's radius plus the height above the earth - in this case, 6.77 x 106 m.

Answer 2

Answer:

The orbital velocity of the smallest possible orbit is 7.91 km/s.

Explanation:

It is given that,

The escape velocity of Earth is 11.2 km/s. We need to find the orbital velocity of the smallest possible orbit. We know that the relation between the escape velocity and the circular orbital velocity.

Escape velocity = √2 × circular orbital velocity

Let v' is the circular orbital velocity. It is given by :

$v'=\dfrac{v}{\sqrt{2} }$

$v'=\dfrac{11.2\ km/s}{\sqrt{2} }$

v' = 7.91 km/s

So, the orbital velocity of the smallest possible orbit is 7.91 km/s. Hence, this is the required solution.

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The escape velocity of earth is 11.2 km/s. Find the orbital velocity of the smallest possible orbit ?

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