Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 3, Problem 42P
To determine

The velocity of a rocket launched at 40.0o North latitude along East, with respect to the ground and the advantage in firing the rocket from a site closer to the equator.

Expert Solution & Answer
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Answer to Problem 42P

The velocity of a rocket launched at 40.0o North latitude along East, with respect to the ground is found to be 1.0645×104 m/s . If the rocket is launched from a point close to the equator, the speed which is to be imparted to it at the ground would beless by 108 m/s , which would save on fuel costs.

Explanation of Solution

Given info:

The latitude of the point of launch, θ=40.0°

Velocity of the rocket with respect to the center of the Earth, vR=11.0 km/s

Formula used:

The velocity of the ground with respect to the ground vG is given by

  vG=2πrT.....(1)

Here, r is the radius of the circular path traversed by the particle and T is the time taken for it to complete one rotation.

Calculation:

The point of launch is at a latitude 40.0o North and as the Earth rotates around its axis, the ground moves in a circular path of radius r ,with its center on the Earth’s axis of rotation.

This is shown in the diagram below:

  Physics Fundamentals, Chapter 3, Problem 42P

From the diagram, it can be seen that,

  r=Rcosθ.....(2)

Here, R is the radius of the Earth, which has a value 6.37×103km .

Express the radius of the Earth in meters.

  R=(6.37× 103km)( 10 3  m 1 km)=6.37×106m

Calculate the value of r by substituting the values of R and θ in equation (2).

  r=Rcosθ=(6.37× 106m)(cos40.0°)=4.88×106m

The Earth takes 24 h to complete one rotation. In the same time, the point on the ground completes its circular path of radius r .

Calculate the time T taken by the point on the ground to go once around the circular path of radius r in seconds.

  T=(24 h)( 3600 s 1 h)=8.64×104s

Substitute the values of r and T in equation (1) and calculate the magnitude of the ground’s velocity relative to the center of the Earth.

  vG=2πrT=2π( 4.88× 10 6 m)( 8.64× 10 4 s)=3.55×102m/s

The Earth rotates from west towards east. Hence, the point where the rocket is launched has a velocity 3.55×102m/s directed eastwards at the instant of launch.

The rocket is launched with a speed of 11.0 km/s eastward with respect to the center of Earth.

Express the speed vR of the rocket with respect to the center of the Earth in m/s.

  vR=(11.0 km/s)( 10 3 m 1 km)=1.10×104m/s

The Ground rotates with a speed vG eastwards with respect to the center of the Earth.

The speed vRG of the rocket with respect to the ground is given by,

  vRG=vRvG

Calculate the speed vRG by substituting the given values in the above equation.

  vRG=vRvG=(1.10× 104m/s)(3.55× 102m/s)=1.0645×104 m/s

If the rocket is launched at a point near the equator, the ground moves in a circular path equal to the radius of the Earth. Calculate the speed of the ground at the equator using equation (1) in the following form:

  (vG)Eq=2πRT

Substitute the given values in the above expression and calculate the velocity of the ground with respect to the center of Earth at equator.

  ( v G)Eq=2πRT=2π( 6.37× 10 6 m)( 8.64× 10 4 s)=4.63×102m/s

The speed that needs to be imparted to the rocket at equator is the relative velocity of the rocket with respect to the ground.

  ( v RG)Eq=vR( v G)Eq=(1.10× 104m/s)(4.63× 102m/s)=1.0537×104 m/s

The difference in the velocities of the rocket when launched at 40.0o latitude and at equator is given by,

  Δv=vRG(v RG)Eq

Substitute the given values in the above expression:

  Δv=vRG( v RG)Eq=(1.0645× 104m/s)(1.0537× 104 m/s)=108 m/s

Conclusion:

Thus, the velocity of a rocket launched at 40.0oNorth latitude along East, with respect to the ground is found to be 1.0645×104 m/s . If the rocket is launched from a point close to the equator, the speed which is to be imparted to it at the ground would be less by 108 m/s , which would save on fuel costs.

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