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Chapter 3: Q. 27P (page 124)

At t = 532.0s after midnight, a spacecraft of mass 1400 kg is located at position (3x105, 7x105,-4x105 ) m, and at that time an asteroid whose mass is 7 x 10 15 kg is located at position (9 x 10 ^ 5, - 3 x 10 ^ 5, - 12 x 10 ^ 5) m. There are no other objects nearby. (a) Calculate the (vector) force acting on the spacecraft. (b) At t=532.0: the spacecraft's momentum was p→iand at the later time t = 538.0s its momentum was p→f.Calculate the (vector) change of momentum p→f-p→i.

Short Answer

Expert verified
  1. The value of vector forceF=7.26×103,0,0
  2. Change in vector momentump→f-p→i=40×103,0,0Kg·m/s

Step by step solution

01

Identification of given data

  • t=532.0s
  • Mass of spacecraftm=1400kg
  • Positions of spacecraft(3×105,7×105,-4×105)
  • Mass of asteroidp→f-p→i=40×103N·m
02

Calculation of the vector force

(a) According to the Newton’s law of gravitation

F=Gm1m2r2

Where

G – Gravitational constant

m1- Mass of 1 object

m2-Mass of 2 objects

r-Distance between two object

F=6.67×10-11×1400×7×10153×1052F=7.26×103NF=⟨7.26×103,0,0⟩N

03

Calculation of change in vector momentum

(b) The spacecraft momentum

When t=532 s

p→i=F·tp→i=7.26×103×532p→i=⟨3.86×106,0,0⟩Kg⋅m/s

When t=538 s

p→f=F·tp→f=7.26×103×538p→f=⟨3.90×106,0,0⟩Kg⋅m/s

Then the value of role="math" localid="1658079511166" p→f-p→i

p→f-p→i=3.86×106-3.90×106p→f-p→i=⟨40×103,0,0⟩Kg⋅m/s

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