# Sputnik I was launched into orbit around Earth in 1957. It had a perigee (the closest approach to Earth, measured from Earth’s center) of 6.81 x 106 m and an apogee (the furthest point from...

Sputnik I was launched into orbit around Earth in 1957. It had a perigee (the closest approach to Earth, measured from Earth’s center) of 6.81 x 106 m and an apogee (the furthest point from Earth's center) of 7.53 x 106 m. What was its speed when it was at its perigee? The mass of Earth is 5.97 x 1024 kg and G = 6.67 x 10-11 N • m2/kg2.

Borys Shumyatskiy | Certified Educator

Hello!

Denote the perigee distance as `r_p` and the apogee distance as `r_a.` The corresponding speeds are `V_p` and `V_a.` Also denote the mass of the Sputnik as `m` and the mass of Earth as `M.` It is true that `m` is negligible compared to `M.`

We'll use two laws of conservation: conservation of angular momentum and conservation of energy.

The first law gives us `r_p V_p = r_a V_a` (note that at apogee and perigee velocity is perpendicular to Sputnik-Earth line).

The second law gives

`(mV_p^2)/2-(mV_a^2)/2=(GMm)/r_p-(GMm)/r_a,` or  `V_p^2/2-(V_a^2)/2=(GM)/r_p-(GM)/r_a.`

Here `(mV^2)/2` is kinetic energy and `-(GMm)/r` is potential energy.

Express `V_a` from the first equation, `V_a=V_p r_p/r_a,` and substitute this into the second equation,

`1/2 V_p^2(1-r_p^2/r_a^2)=GM(1/r_a-1/r_b).`

Simplifying obtain `V_p^2=2GM 1/(r_a+r_p) r_a/r_p,` so

`V_p=sqrt(2GM 1/(r_a+r_p) r_a/r_p).` Also `V_a=sqrt(2GM 1/(r_a+r_p) r_p/r_a).`

So we need the mass of Earth multiplied by the gravitational constant `G.` Google says it is about `4* 10^14 m^3 / s^2.`

This way the result in numbers is

`V_p=sqrt(2*4* 10^14 *1/(6.81+7.53)*10^(-6)*7.53/6.81) approx 7800(m/s).`

electreto05 | Certified Educator

During its movement, the satellite is subjected to a centrifugal force, pointing towards the outside of the trayectory. In addition, is subjected to the force of Earth's gravity.

To keep the satellite in orbit, the speed must be such that, the centrifugal force it is equal and opposite to the force of gravity; so we can consider the following equality:

Fc = Fg

(m*v^2)/r = (G*M*m)/r^2

Where:

M,  is the mass of the earth.

m,  is the mass of the satellite.

r,  is the distance between the satellite and the earth

G,  is the gravitational constant.

v,  is the tangential velocity of satellite.

Solving, for the speed at perigee:

v^2 = (G*M)/r

v = sqrt (GM/r) = sqrt [(6.67*10-11)(5.97*10^24)/(6.81*10^6)]

v = 7.64*10^3 m/s  =  7.64 km/s

So, at perigee the satellite must have a speed of 7.64 km/s