False the North Star never changes it position
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Answer:
3 m/s
Explanation:
We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:
Initial displacement (d₁) = 4 m
Final displacement (d₂) = 16 m
Change in displacement (Δd) =?
Δd = d₂ – d₁
Δd = 16 – 4
Δd = 12 m
Finally, we shall determine the determine the average velocity. This can be obtained as follow:
Change in displacement (Δd) = 12 m
Time (t) = 4 s
Velocity (v) =?
v = Δd / t
v = 12 / 4
v = 3 m/s
Thus, the average velocity of the jogger is 3 m/s
In both magnitude and direction since acceleration is a vector quantity
