What velocity is necessary to place a T.A.R.D.I.S. in a stable orbit 30,000 km above the surface of the Earth? (Considering gravitation) If the...

Since your question contains multiple parts, I will provide a detailed answer to part A and some guidance for parts B and C.

A)When the satellite is in a stable orbit, its speed is constant. The centripetal acceleration is equal `v^2/R` , where R is the radius of the orbit, and is provided by the gravitational force between the satellite and the Earth: `(GMm)/R^2` , where M and m are the masses of the Earth and the satellite, respectively, and G is the gravitational constant.

According to the second Newton's Law

`(GMm)/R^2 = m v^2/R`

From here, the speed of the satellite is

`v = sqrt((GM)/R)`

The values of G and M are given, and the radius of the orbit is the sum of the radius of the Earth and the altitude of the satellite:

`R = 6.371*10^6 + 3*10^7 = 36.371*10^6` m, or approximately `3.6*10^7` m.

So the speed needed to launch the satellite in the orbit with this radius is

`v = sqrt((6.674*10^(-11)*(5.972*10^24))/(3.6*10^7)) = 3.33*10^3 m/s`

B) The work that the engines of the satellite must exert in order to increase the altitude can be calculated as the difference of the values of the potential energy of the satellite at these altitude:

`W = U_1 - U_2 = (GMm)/R_1 - (GMm)/R_2` . Here, `R_1` is the radius of the original orbit calculated in part A, and `R_2` is the radius of the new orbit that can be calculated the same way, by adding the radius of the Earth to the new altitude.

C) The velocity of the satellite at the new altitude can be calculated using the same result as in part A, but for the radius of the new orbit:

`v = sqrt((GM)/R_2)` .

Hope this helps :)