a satellite orbiting Earth at an orbital radius r has a velocity v. What would the velocity be if the satellite is moved to an orbital radius of 6r?
sorry.
The speed is inversely proportional to the square root of the orbit radius
so, v/√6
The velocity of a satellite in orbit around Earth can be determined using Kepler's Third Law and the law of conservation of angular momentum.
Kepler's Third Law states that the square of the orbital period (T) of a satellite is proportional to the cube of its orbital radius (r). Mathematically, it can be expressed as:
T^2 ∝ r^3
If the satellite is moved to a new orbital radius of 6r, then the new orbital period (T') can be calculated as follows:
T'^2 ∝ (6r)^3
T'^2 ∝ 216r^3
Since the orbital period is directly related to the velocity of the satellite, we can express the proportional relationship as:
T'^2 = k * r^3
where k is a constant.
Since the satellite is still orbiting Earth, we can assume that the gravitational force acting on it remains the same. The law of conservation of angular momentum states that the product of the moment of inertia (I) and the angular velocity (ω) of the satellite remains constant. In a circular orbit, the moment of inertia is proportional to the square of the orbital radius (I ∝ r^2) and the angular velocity is inversely proportional to the orbital radius (ω ∝ 1/r). Thus, we can express this relationship as:
I * ω = k' * r^2 * (1/r) = k'' * r
where k' and k'' are constants.
Since velocity (v) is equal to the product of the angular velocity (ω) and the orbital radius (r), we can rewrite the equation as:
v = k'' * r
Therefore, if the satellite is moved to an orbital radius of 6r, the velocity (v') at the new orbital radius can be calculated as follows:
v' = k'' * (6r)
v' = 6 * (k'' * r)
v' = 6v
Thus, the velocity of the satellite would be 6 times greater if it is moved to an orbital radius of 6r.
To find the velocity of a satellite when moved to a new orbital radius, we can use the principle of conservation of angular momentum. According to this principle, the product of the moment of inertia and angular velocity of a rotating body remains constant unless acted upon by external torques.
In the case of a satellite orbiting around the Earth, the angular momentum is conserved as it moves to a new orbital radius. Mathematically, this can be expressed as:
I₁ * ω₁ = I₂ * ω₂
Where:
I₁ and I₂ are the initial and final moments of inertia respectively.
ω₁ and ω₂ are the initial and final angular velocities respectively.
In this case, since the mass of the satellite remains constant, the moment of inertia is directly proportional to the square of the orbital radius (I ∝ r²).
I₁ * ω₁ = I₂ * ω₂
(m₁ * r₁²) * ω₁ = (m₂ * r₂²) * ω₂
Considering that mass (m) is also constant, it cancels out from the equation, simplifying it to:
r₁² * ω₁ = r₂² * ω₂
Now, we can compare the initial and final orbital radii. Let's say the initial orbital radius is denoted by r₁, and the final orbital radius is 6 times the initial radius, i.e., 6r₁.
Plugging in these values into the equation, we get:
(r₁)² * ω₁ = (6r₁)² * ω₂
r₁² * ω₁ = 36r₁² * ω₂
ω₂ = ω₁ / 36
This equation tells us that the final angular velocity (ω₂) is 1/36th of the initial angular velocity (ω₁).
The velocity of the satellite is given by the product of the orbital radius and the angular velocity:
v = r * ω
Substituting the value of ω₂ into this equation, we get:
v₂ = (6r₁) * (ω₁ / 36)
v₂ = (r₁ / 6) * ω₁
Therefore, the velocity of the satellite when moved to an orbital radius of 6r is given by v₂ = (r₁ / 6) * ω₁.