An object of mass 3.5x1023kg circles the earth and is attracted to it with a force whose magnitude is given be GMeM/r^2. If the period of rotation is 22 days, what is the distance from the earth to the object? Here G=6.67x10-11 Nm2/kg2, Me=6x1024 kg.
G Me M/r^2 = M V^2/r
Express V in terms of the period P = 22 days = 1.90*10^6 seconds
2 pi r/P = V
G Me/r^2 = 4 pi^2 r/P^2
4 pi^2/(G*Me) = r^3/P^2
Solve for r.
To find the distance from the Earth to the object, we need to use the formula for the period of rotation of a satellite around a central planet:
T = 2π√(r^3 / GM)
Where:
T is the period of rotation (in seconds)
r is the distance from the object to the center of the Earth
G is the gravitational constant (6.67x10^-11 Nm^2/kg^2)
Me is the mass of the Earth (6x10^24 kg)
We first need to convert the given period of rotation from days to seconds. There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute.
22 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute = 1,900,800 seconds
Now, let's plug in the known values into the formula for the period of rotation:
1,900,800 seconds = 2π√(r^3 / (6.67x10^-11 Nm^2/kg^2)(6x10^24 kg))
Simplifying:
1,900,800 seconds = 2π√(r^3 / (4.002x10^29 Nm^2)
Next, we square both sides of the equation to eliminate the square root:
(1,900,800 seconds)^2 = 4π^2(r^3 / (4.002x10^29 Nm^2)
Simplifying further:
3.617x10^12 seconds^2 = π^2(r^3 / (1.001x10^29 Nm^2)
Now, rearrange the equation to solve for r:
r^3 = (3.617x10^12 seconds^2 x 1.001x10^29 Nm^2) / π^2
Take the cube root of both sides to isolate r:
r = ∛((3.617x10^12 seconds^2 x 1.001x10^29 Nm^2) / π^2)
Calculating this expression will give you the distance from the Earth to the object.