A satellite orbits the Earth (mass = 5.98 x 1024 kg) once every = 43200 s. At what radius does the satellite orbit?

26621 km

Explanation:

We are given;

Mass: m = 5.98 x 10^(24) kg

Period; T = 43200 s

Formula for The velocity(v) of the satellite is:

v = 2πR/T

Where R is the radius

Formula for centripetal acceleration is;

a_c = v²/R

Thus; a_c = (2πR/T)²/R = 4π²R/T²

Formula for gravitational acceleration is:

a_g = Gm/R²

Where G is gravitational constant = 6.674 × 10^(-11) m³/kg.s²

Now the centripetal acceleration of the satellite is caused by its gravitational acceleration. Thus;

Centripetal acceleration = gravitational acceleration.

Thus;

4π²R/T² = Gm/R²

Making R the subject gives;

R = ∛(GmT²/4π²)

Plugging in the relevant values;

R = ∛((6.674 × 10^(-11) × 5.98 x 10^(24) × 43200²)/(4 × π²))

R = 26.621 × 10^(6) m

Converting to km, we have;

R = 26621 km

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s