Answer:
The correct option is;
B) 2.2 hr
Explanation:
The location of the rock in the orbit of the asteroid = Close to the surface of the asteroid
The required equation is given as follows;
[tex]\dfrac{G \times M \times m}{r^2} = \dfrac{m \times v^2}{r}[/tex]
[tex]\therefore v^2 = \dfrac{G \times M}{r}[/tex]
Where;
v = The orbital velocity
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r = The radius of the planet = 16 km = 16,000 m
M = The mass of the planet = 4 × 10¹⁶ kg
∴ v² = 6.67430 × 10⁻¹¹ × 4 × 10¹⁶/16,000 = 166.8575
v = √(166.8575) = 12.92 m/s
The orbital velocity = v = 12.92 m/s
The time it takes for the rock to make one orbit round the asteroid is given as follows
The length of the orbit = The circumference of the asteroid = 2 × π × r
The length of the orbit = 2 × π × 16,000 ≈ 100530.965 m
The time it takes for the rock to make one orbit round the asteroid = The length of the orbit/(The speed of the asteroid)
The time it takes for the rock to make one orbit round the asteroid = 100530.965/(12.92) ≈ 7781.034 seconds or 2.2 hours (by approximation by one decimal place)
The time it takes for the rock to make one orbit round the asteroid ≈ 2.2 hours.