A 17.0-kg satellite has a circular orbit with a period of 2.31 h and a radius of 9.10×106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 8.10 m/s2, what is the radius of the planet?
radius = m
figure the centripetal acceleration v^2/r
where v=2PIr/T you are given r as 9.10E6
now set that acceleration equal to
8.10*(rp/9.1E6)^2 and solve for the radius of the planet rp.
What is Plr?
Which values are solutions to the inequality below? Check all that apply. sqrtx
To find the radius of the planet, we can use Newton's law of universal gravitation, which states that the force of gravity between two objects is given by:
F = (G * m1 * m2) / r^2
Where:
F is the force of gravity
G is the gravitational constant (6.67430 × 10^-11 N m^2/kg^2)
m1 and m2 are the masses of the two objects
r is the distance between the centers of the two objects
In this case, the satellite is in a circular orbit around the planet, so the gravitational force is providing the necessary centripetal force to keep the satellite in orbit. Therefore, we have:
F = (m * v^2) / r
Where:
m is the mass of the satellite
v is the orbital velocity of the satellite
r is the radius of the orbit
We can express the orbital velocity in terms of the period (T) of the satellite's orbit and the radius (r) of the orbit using the formula:
v = (2 * π * r) / T
Substituting this into our previous equation, we have:
F = (m * ((2 * π * r) / T)^2) / r
As the magnitude of the gravitational acceleration on the surface of the planet is given as 8.10 m/s^2, we can equate this to:
F = (G * m * M) / r^2
Where:
M is the mass of the planet
Now we can equate the two expressions for F:
(m * ((2 * π * r) / T)^2) / r = (G * m * M) / r^2
Canceling out the mass of the satellite (m) and simplifying, we get:
((2 * π * r) / T)^2 = (G * M) / r
Simplifying further:
(2 * π * r)^2 / T^2 = (G * M) / r
Cross-multiplying and rearranging the equation, we get:
r^3 = (G * M * T^2) / (4 * π^2)
Finally, solving for the radius of the planet (r), we have:
r = cube root of ((G * M * T^2) / (4 * π^2))
Now, we can plug in the given values:
- m = 17.0 kg (mass of the satellite)
- T = 2.31 hours = 2.31 * 3600 seconds (period of the satellite's orbit)
- G = 6.67430 × 10^-11 N m^2/kg^2 (gravitational constant)
- g = 8.10 m/s^2 (gravitational acceleration on the surface of the planet)
Let's calculate the radius of the planet: