Question

If a satellite orbiting the Earth has a period of 125 min, then it must be 1.90 x 106 m above the Earth’s surface. Note: The mass of earth is 5.97 x 1024 kg, and the mean radius of earth is 6.38 x106 m.

Answer 1

t Stable orbit problem.

The satellite has negligible mass compared to the earth, so this is a theoretical (single body) problem ignoring its mass altogether.

(The maths is a lot less than a two bodyproblem)

Note : each orbital radius has only one stable orbit time, and vice versa.

(Same planet mass assumed)

So, you've defined the (sidereal) orbit time (t) @ 125 minutes ( 7,500 seconds ), and you want the satellites orbital radius (r) ?

First calculate the rotation rate (w) of the satellite = ( 2 * pi ) / 7,500 = 8.37758 e-4 rad / sec

The equation build is as follows :

For stable orbit, gravitational accelearation (g) = centripetal acceleration (a)

So : ( G * m ) / r ² = v ² / r

Transpose to feature v : v = square root ( ( G * m ) / r )

(keep for later)

Bring in standard equation : w = v / r, transpose to feature r : r = v / w ,

(bring back : v = square root ( ( G * m ) / r ) )

So : r = ( square root ( ( G * m ) / r ) ) / w

Traspose to : r = cube root ( ( G * m ) / w ² )

r = 8.2798 e6 meters (orbital radius)

So, the distance above the earths surface = r - R = 1.9001 e6 meters

please mark me as brainliest

Answer 2

Answer:

T = 2pi * rad(r^3 / G * m)

125 = 2pi * rad((d + 6.38 * 10^6)^3 / 6.67*10^-11 * 5.97 *10^24)

7,500 = 2pi * rad((d + 6.38 * 10^6)^3 / 6.67*10^-11 * 5.97 *10^24)

1,193.662073 = rad((d + 6.38 * 10^6)^3 / 6.67*10^-11 * 5.97 *10^24)

1,424,829.145 = (d + 6.38 * 10^6)^3 / 6.67*10^-11 * 5.97 *10^24

1,424,892.145 = (d + 6.38 * 10^6)^3 / 3.98199 *10^14

5.673906272 * 10^20 = (d + 6.38 * 10^6)^3

cube root(5.673906272 * 10^20) = (d + 6.38 * 10^6)

8278672.817= d + 6.38 * 10^6

d = 1.90 *10^6 m

Explanation: solve it using Kepler's 3rd Law