MasteringPhysics: chapter 12 - Google Chrome Secure l https://session.masteringp

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MasteringPhysics: chapter 12 - Google Chrome Secure l https://session.masteringphysics.com/myct/itemView?offset-next&assignmentProblemID;=84003509 chapter 12 Problem 12.52 2.5Ws Resources previous 3 of 10 | next Problem 12.52 Part A At a certain instant, the earth, the moon, and a stationary 1300 kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 × 105 km in length. Find the magnitude of the net gravitational force exerted on the spacecraft by the earth and moon. Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining, no points deducted Part B Find the direction of the net gravitational force exerted on the spacecraft by the earth and moon State the direction as an angle measured from a line connecting the earth and the spacecraft Submit My Answers Give Up Part C What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun. Submit My Answers Give Up

Explanation / Answer

a)

we know,

M_earth = 5.98*10^24 kg
M_moon = 7.35*10^22 kg

m = 1300 kg

r = 3.84*10^8 m

F_earth = G*M_earth*m/r^2

= 6.67*10^-11*5.98*10^24*1300/(3.84*10^8)^2

= 3.51 N

F_moon = G*M_moon*m/r^2

= 6.67*10^-11*7.35*10^22*1300/(3.84*10^8)^2

= 0.0432 N

Fnet = sqrt(F_eath^2 + F_moon^2 + 2*F_earth*F_moon*cos(60))

= sqrt(3.51^2 + 0.0432^2 + 2*3.51*0.0432*cos(60))

= 3.53 N <<<<<<<<---------Answer

B)

tan(theta) = 0.0432*sin(60)/(3.51 + 0.0432*cos(60))

theta = tan^-1(0.0106)

= 0.607 degrees <<<<<<<<---------Answer

3)
Workdone = change in gravitational potential energy

= Uf - Ui

= 0 - ( -G*M_earth*m/r - G*M_moon*m/r)

= 6.67*10^-11*5.98*10^24*1300/(3.84*10^8) + 6.67*10^-11*7.35*10^22*1300/(3.84*10^8)

= 1.37*10^9 J

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