An artificial satellite circles the earth in a circular orbit at a location where the acceleration due to gravity is 6.03 m/s^2. determine the orbital period of the satellite.

Orbital period, T = 2.02 hours

Explanation:

It is given that, an artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.03 m/s². We have to find the orbital period (T) of the satellite.

Firstly, calculating the distance between Earth and satellite. The acceleration due to gravity is given by :

G = universal gravitational constant

M = mass of earth

r = 8126273.3 m..........(1)

Now, according to Kepler's third law :

Putting the value of r from equation (1) in above equation as :

T = 7294.01 seconds

Since, 1 hour = 3600 seconds

Converting seconds to hour we get :

So, T = 2.02 hour  

So, the orbital period of the satellite is 2.02 hours.

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .