Answer:
2917.4 m/s
Explanation:
From the question given above, the following data were:
Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth
Radius (r) of the Moon = 1737 Km
Escape velocity (v) =?
Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:
Gravitational acceleration of the earth = 9.8 m/s²
Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth
= 0.25 × 9.8 = 2.45 m/s²
Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:
1 Km = 1000 m
Therefore,
1737 Km = 1737 Km × 1000 m / 1 Km
1737 Km = 1737000 m
Thus, 1737 Km is equivalent to 1737000 m
Finally, we shall determine the escape velocity of the rocket as shown below:
Gravitational acceleration of the Moon (g) = 2.45 m/s²
Radius (r) of the moon = 1737000 m
Escape velocity (v) =?
v = √2gr
v = √(2 × 2.45 × 1737000)
v = √8511300
v = 2917.4 m/s
Thus, the escape velocity is 2917.4 m/s
