Asatellite is orbiting at a height of about 650 km above the earth's surface. the mass of the earth is 5.976 x 10^24 kg and the radius of the earth is 6.378 x 10^6 m. a) assuming a circular orbit, what is the speed of the satellite in its orbit? b) assuming a circular orbit, what is the period of the satellite's orbit?

a) v = 7533.3 m/s

b) T = 54677.0 seconds

Explanation:

let Fc be the centripetal force and Fg be the gravitational force.

m be the mass of the satelite and M be the mass of the earth.

a) A satelite in orbit experiences two forces, centripetal force and gravitation force due to the center of mass.

this forces are balanced, meaning at each point in the orbit:

Fc = Fg

m×v^2/r = G×m×M/(r^2)

because r is the same and m is the same:

v^2 = G×M/r

      = [(6.67408×10^-11)×(5.976×10^24)]/(6378×10^6 + 650×10^3)

      = 56750572.11

then v = \sqrt{56750572.11} = 7533.3 m/s.

therefore, the speed of the satelite in it's orbit will be 7533.3 m/s.

b)  for a given circular orbit, the satelite will cover a displacement of 2πr in speed of 2πr/T where T is the period of the orbit:

then.

2πr/T = \sqrt{G×M/r} = v

T = 2πr/(v)

  = 2π(6378×10^6 + 650×10^3)/(7533.3)

  = 54677.0 seconds

therefore, it will take the satelite 54677 secords to orbit earth.

answer; (d) spontaneous processes are characterized by the overall conversion of order to disorder;

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.