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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: D
Rs = 10.0 m/s
The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s
Explanation:
Since the boat is moving perpendicular to the current of the river, the speed of the boat has two components.
i. 8.0m/s in the direction perpendicular to the current
ii. 6.0m/s in the direction of the current.
So, the resultant speed can be derived by using the equation;
Rs = √(Rx^2 + Ry^2)
Taking
Ry = 8.0m/s
Rx = 6.0m/s
Substituting into the equation, we have;
Rs = √(6.0^2 + 8.0^2)
Rs = √(36+64) = √100
Rs = 10.0 m/s
The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s
