a satellite orbiting earth at an orbital radius r has a velocity v which represents the velocity if the satellite is moved to an orbital radius of 5r?
1/√5 v
1/5v
√5v
5v
The velocity v of a satellite in orbit is determined by the balance between the gravitational force pulling the satellite inward and the centrifugal force pushing it outward. This balance can be described by the equation:
F_gravity = F_centrifugal
Using the equation for gravitational force:
G * (m_earth * m_satellite) / r^2 = (m_satellite * v^2) / r
Here G is the gravitational constant, m_earth is the mass of the Earth, m_satellite is the mass of the satellite, and r is the orbital radius.
If we rearrange this equation to solve for v, we get:
v^2 = (G * m_earth) / r
Taking the square root of both sides, we find:
v = √(G * m_earth / r)
Now we can determine the velocity v' if the satellite is moved to an orbital radius of 5r. Let's denote v' as the new velocity. Using the formula above with r replaced by 5r, we find:
v' = √(G * m_earth / (5r))
Simplifying this expression, we get:
v' = 1/√5 * √(G * m_earth / r)
Therefore, the velocity of the satellite if moved to an orbital radius of 5r is 1/√5 v.
Which form of Kepler's third law can you use to relate the period T and radius R of a planet in our solar system as long as the unit year is used for the period and astronomical unit is used for the radius?
T^2 = r^3
T^2=r
T= r^2
T^3 = r^2
The correct form of Kepler's third law that relates the period T and radius R of a planet in our solar system is:
T^2 = r^3
Here, T represents the period in years and r represents the radius in astronomical units.
The mass of an object doubles. What happens to the gravitational force between it and another object who's mass stays the same, assuming they remain the same distance apart?
The amount of the gravitational force decreases by one half.
The amount of gravitational force doubles.
The amount of gravitational force quadruples.
The amount of the gravitational force remains the same.
The amount of gravitational force remains the same.
According to Newton's law of universal gravitation, the gravitational force between two objects is directly proportional to the product of their masses. If one object's mass doubles while the other object's mass remains the same, the overall gravitational force between them will still be doubled. Therefore, the amount of gravitational force remains the same.
Which are the features of Copernicus's model of the solar system? Select THREE correct answers.
The outer planets move in epicycles.
The stars are fixed to an outer sphere.
The planets move in perfect circles.
The sun is at the center.
All planets orbit the Sun at the same speed.
Earth is fixed to a crystalline sphere.
The three correct features of Copernicus's model of the solar system are:
1. The planets move in perfect circles.
2. The sun is at the center.
3. All planets orbit the Sun at the same speed.
for the moon orbiting it's planet, r p is the shortest distance between the moon and it's planet R a is the longest distance between the moon and it's planet. What is the moons orbital eccentricity if r p is equal to 0.27r a?
0.65
0.48
0.27
0.57
The eccentricity of an orbit can be calculated using the formula:
eccentricity = (R_a - r_p) / (R_a + r_p)
Given that r_p is equal to 0.27r_a, we can substitute the value in the formula:
eccentricity = (R_a - 0.27R_a) / (R_a + 0.27R_a)
Simplifying the expression:
eccentricity = 0.73R_a / 1.27R_a
The R_a terms cancel out, resulting in:
eccentricity = 0.73 / 1.27 ≈ 0.57
Therefore, the moon's orbital eccentricity is approximately 0.57.
A line between a planet and the sun sweeps out two equal areas at different places as it moves along it's orbit. Which factor remains constant as this happens?
The distance from the sun to the planet while it sweeps out the areas.
The time require to sweep out the areas.
The distance the planet moves along it's orbit while it sweeps out the area.
The planets speed as it sweeps out the areas.