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Alina [70]
3 years ago
6

(b) An example of an inverse square field is the gravitational field F = −(mMG)r/|r|3. Use part (a) to find the work done by the

gravitational field when the earth moves from aphelion (at a maximum distance of 1.52 108 km from the sun) to perihelion (at a minimum distance of 1.47 108 km). Use the values m = 5.97 1024 kg, M = 1.99 1030 kg, and G = 6.67 ✕ 10−11 N·m2/kg2. (Round your answer to two decimal places.)
Physics
1 answer:
enyata [817]3 years ago
4 0

Answer:

To solve this, we need to recall that Work done W,

W = F x dₐ                         Eqn 1

where F is the gravitational field force and d is the distance travelled by the earth.

But F = −(mMG)r/|r|³                    Eqn 2

then, W = −∫ (mMG)r/|r|³ x dₐ

Integrating this expression gives us

W = -4mMG ((1/d1) -(1/d2))

m = 5.971024 kg, M = 1.991030 kg G = 6.67 ✕ 10⁻¹¹ N·m²/kg²

d2 = 1.52x10¹¹m, d1 = 1.47x10¹¹m  make the units in S.I units and similar

Substituting these value in to the eqn above gives

Now , W = 1.77 x 10³²J  (2 Decimal Places)

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An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

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Answer:

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Explanation:

Potential energy is defined as the product of mass by gravity by height.

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