Answer:
a) v = 4.08 m / s , b) T = 2.58 h
Explanation:
For this exercise, let's start by using Newton's second law where force is the attraction of gravity
F = m a
Force is
F = G m [tex]M_{D}[/tex] / r²
The acceleration is centripetal
a = v² / r
Let's replace
G m [tex]M_{D}[/tex] / r² = m v² / r
G [tex]M_{D}[/tex] / r = v²
v = √ G [tex]M_{D}[/tex] / r
The radius is half the diameter
r = d / 2
r = 12 10³/2 = 6.0 10³ m
v = √ (6.67 10⁻¹¹ 1.5 10¹⁵/6 10³
v = √ 16,675
v = 4.08 m / s
If you throw the ball with this speed, take a full turn
c) The constant speed module only changes the direction, so we can use the relationship
v = d / t
The time is
t = d / v
The distance along the circle and this time is called Period
d = 2π r
d = 2π 6 10³
d = 37,70 10³ m
Let's calculate
T = 37.70 10³ / 4.08
T = 9.24 10³ s
Let's reduce to hours
T = 9.24 10³ s (1h / 3600 s)
T = 2.58 h