1.The moon orbits 384400km above the surface of the earth. a).What is the moon velocity ? b).At this velocity, how long does it take the moon to orbit the earth? c).What is the strength of the earth gravitational force at this orbital distance?

1 Answer
Feb 18, 2018

Given that

  • the height of moon's orbit from earth #h=384400km#

  • radius of earth #R_e=6371km#

  • so radius of moon's orbit #R_o=R_e+h=390771*10^3m#

By Newton's law of gravitation the force of gravitational pull on moon is given by

#F_e=(GMm)/R_o^2#,

Where

  • #M-># mass of earth

  • #m-># mass of moon

  • #G-># Gravitational constant.

#F_e# will provide the centripetal force required for the circular orbital motion of the moon.
If the uniform orbital angular velocity of moon is represented by #omega# then

#momega^2R_o=(GMm)/R_o^2#

#=>omega^2=(GM)/R_o^3#

#=>omega^2=(gR_e^2)/R_o^3# #color(red)([" inserting "GM=gR_e^2])#
Where #g=9.8ms^-2# is the acceleration due to gravity at earth's surface.

#omega=R_e/R_o xxsqrt(g/R_o)#

(a) So orbital velocity of moon

#V_m=omega*R_o=Re xx sqrt(g/R_o)#

#V_m=(6371*10^3)*sqrt(9.8/(390771*10^3))~~1008.9ms^-1#

(b) Now time period of rotation of moon will be

#T=(2pi)/omega=(2pi R_o^(3/2))/(R_exxsqrtg)~~28.17days#

(c)The strength of the earth gravitational force at this orbital distance will be represented by the acceleration due to gravity #g'# in moon's orbit

#g'=F_e/m=(GM)/R_o^2=(gR_e^2)/R_o^2#

#=9.8xx(6371*10^3)^2/(390771*10^3)^2=2.6*10^-3ms^-2#