Please solve q 82?

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2 Answers
Apr 3, 2018

Answer:

The value of #g# would change almost not at all. There is a small reduction in the downward force acting on a body due to the rotation of Earth, but its magnitude is far smaller than the gravitational force at the surface.

Explanation:

In response to this question, I calculated the answer to this question, if we were at the equator, where the change would be greatest: https://socratic.org/questions/5667107b7c01493f09503ec0?source=search

It turns out that the rotational motion's contribution is about #1/3000# that of the gravitational field. I maintain that that means there is essentially no change.

Apr 3, 2018

Answer:

#"(3) increases"#

Explanation:

Acceleration due to gravity at a latitude (#phi#) is given as

#"g"_(phi) = "g"_"e" - "R"ω^2cos^2(phi)#

where

  • #"g"_"e" =#Acceleration due to gravity at equator
  • #"R ="# Radius of Earth
  • #ω =#Angular Velocity with which earth is rotating
  • #phi =#Angle between the equatorial plane and the line joining the point where acceleration due to gravity is #"g"_phi# to the centre of the earth. It ranges from #0°# at equator to #90°# at poles.

When Earth was rotating

#"g"_1 = "g"_"e" - "R"ω^2cos(phi)#

Note: #cos^2(phi)# can’t be negative. As #phi# ranges from #0°# to #90°# only.

When Earth stops rotating #(ω = 0)#

#"g"_2 = "g"_"e" - 0#

#"g"_2 = "g"_"e"#

In second case we are not subtracting any term from #"g"_"e"#. Therefore, #"g"_2 > "g"_1#. Acceleration is increased.

If you’re interested to know the derivation of that equation, click here.