On 16th August 2020, an asteroid (later called 2020 QG) has been recorded as the closest asteroid (spotted so far) that flew by the Earth without colliding with it. At the nearest point
of approach, it was only 2 950 km above Earth’s surface and it had a velocity v = 12.3 km·s
−1
.
How much higher was its velocity compared to the escape velocity at that height above Earth’s
surface? Find the ratio v/vesc.

V/Ve = 1.1

Explanation:

The formula to find the escape velocity of an object at a distance from Earth's Surface is given as follows:

Ve = √[2GM/R]

where,

Ve = escape velocity = ?

G = Universal Gravitational  Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

R = Radius of Earth = 6.37 x 10⁶ m

Therefore,

Ve = √[2(6.67 x 10⁻¹¹ N.m²/kg²)(5.97 x 10²⁴ kg)/(6.37 x 10⁶ m)]

Ve = 11.2 x 10³ m/s = 11.2 km/s

Hence, the ratio will be:

V/Ve = (12.3 km/s)/(11.2 km/s)

V/Ve = 1.1

the answer would be   dignified - apex