A1 036-kg satellite orbits the earth at a constant altitude of 98-km. (a) how much energy must be added to the system to move the satellite into a circular orbit with altitude 198 km? mj (b) what is the change in the system's kinetic energy? mj (c) what is the change in the system's potential energy?

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = = 98 km = 98000 m

∴Initial Energy ()  = m +

Substituting v= in the above equation and simplifying we get,

=

Similarly for final condition,

h= = 198km = 198000 m

∴Final Energy() =

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = -

        = ( - )

Substituting ,

M = 6 × kg

m = 1036 kg

G = 6.67 ×

R = 6400000 m

= 98000 m

= 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = m[ - ]

                                                          = [ - ]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[ - ]

                                                             = 2ΔE

                                                             = 968.907 MJ

measuring things 1 litre of water

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2