1 point14. Which factor is not needed hen calculating the velocity of a satellite
orbiting a planet?
the mass of the satellite
the orbital radius of the satellite
the mass of the planet
the universal gravitational constant
Pls help I will give extra points
Answers
The mass of the satellite is not required when calculating the velocity of a satellite orbiting a planet.
Given that the centripetal force on the satellite is;
F = mv^2/r
Where;
F = centripetal force that keeps the satellite in its orbit
m = mass of the satellite
r = radius of the satellite
Since the force of gravity and the centripetal force both act on the satellite and they are exactly balanced;
F = GMm/r^2
Where;
G = gravitational constant
M = mass of the planet
m = mass of the satellite
r = radius of the satellite
Hence;
F = GMm/r^2 = mv^2/r
GMm/r^2 = mv^2/r
v = √GM/r
Thus, the mass of the satellite is not required when calculating the velocity of a satellite orbiting a planet.
Learn more: brainly.com/question/21454806
Answer:
i. the mass of the satellite will not be required in this calculation
Explanation:
When a satellite is orbiting a planet, it experiences two forces. The centripetal force and the gravitational force that the planet exerts on the satellite. In order for the satellite to keep in orbit, the centripetal force and the gravitational force must be equal.
The expression for the centripetal force is:
F_c = (m_s)v² / R
where
m_s is the mass of the satellite
R is the radius of the satellite's orbit
v is the velocity that the satellite travels with around the planet
The expression for the Gravitational force is:
F_g = (G M_p m_s) / R²
where
G is the universal gravitational constant
M_p is the mass of the planet
m_s is the mass of the satellite
R is the radius of the satellite's orbit
Thus, equating the two forces together, we get:
(G M_p m_s) / R² = (m_s)v² / R
We can cancel out m_s since it is a common factor on both sides.
Thus,
(G M_p) / R² = v² / R ⇒ M_p = v²R / G
Therefore, the mass of the satellite is not required to calculate the mass of the planet.
Explanation:
Hope this helps:)
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