What is the kinetic energy of a 200 kg satellite as it follows a circular orbit of radius 8x106m around the earth?

Kinetic energy of satellite will be 4.98 × 10^9 J.

Explanation:

The velocity V of a satellite orbiting around the earth is given by  

                            V = √(GM/r)    ………. (i)

Where,  

G = gravitational constant = 6.67 × 10^-11 Nm^2/kg^2

M = mass of earth = 5.97 × 10^24 kg

r = radius of orbit = 8 × 10^6 m

By putting values in  equation (i),

                              V = √((6.67 × 10^-11)(5.97 × 10^24)/ 8 × 10^6)    

                              V = 7055.13 m/s

We know that,  

                            Kinetic energy = ½ mV^2

Where,

m = mass of satellite  

so,

                           Kinetic energy = ½ (200)(7055.13)^2

                           Kinetic energy = 4.98 × 10^9 J

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.