1. At what distance above the Earth (mass of 5.97 x 1024 kg and radius of 6.38 x 106 m) would a satellite have a period of 125 min?
F = G m M/r^2
Ac = F/m = G M/r^2 = v^2/r
so
G M = r v^2
125 min = 7500 s
v * 7500 = 2 pi r
v = pi r/3750
v^2 = pi^2 r^2/1.4*10^7
so
G M = 9.87 r^3 /1.4*10^7
but G = 6.67*10^-11 and M is 5.97*10^24
so
r^3 =(1.4*10^7/9.87)(6.67*10^-11*5.97*10^24)
height above earth = r - 6.38*10^6
To find the distance above the Earth at which a satellite would have a period of 125 minutes, we can use the formula for the orbital period of a satellite.
The orbital period of a satellite is given by:
T = 2π√(r³/GM)
Where T is the period, r is the distance from the center of the Earth to the satellite, G is the gravitational constant (approximately 6.67430 x 10^-11 m^3/(kg s^2)), and M is the mass of the Earth.
In this case, we are given the period T as 125 minutes, which we need to convert to seconds:
T = 125 min * 60 s/min = 7500 s
Given the values of the mass of the Earth (M = 5.97 x 10^24 kg) and the radius of the Earth (r = 6.38 x 10^6 m), we can substitute these values into the equation and solve for the distance r.
7500 s = 2π√(r³/(6.67430 x 10^-11 * 5.97 x 10^24))
Simplifying the equation:
7500 s = 2π√(r³/(3.98950 x 10^15))
Now, isolate the cube of the distance:
(r³/(3.98950 x 10^15)) = (7500 s / 2π)²
Multiply both sides by (3.98950 x 10^15) to get rid of the denominator:
r³ = (7500 s / 2π)² * (3.98950 x 10^15)
Finally, take the cube root of both sides to solve for r:
r = ∛[(7500 s / 2π)² * (3.98950 x 10^15)]
Using a calculator, compute the value of the expression inside the cube root, and then take the cube root to find the value of r. This will give you the distance above the Earth at which a satellite would have a period of 125 minutes.
don't you have a handy formula relating mass and radius to orbital period?
Just plug in your numbers. Watch the units.