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What is the approximate tangential speed of an object orbiting earth with a radius of 1.8 × 108 m and a period of 2.2 × 104 s? 7.7 × 10–4 m/s 5.1 × 104 m/s 7.7 × 104 m/s 5.1 × 105 m/s

Respuesta :

hamzaahmeds

This question involves the concepts of time period and tangential speed.

The tangential speed of the object is "5.1 x 10⁴ m/s".

What is the time period?

The following formula gives the time period of the object:

[tex]T=\frac{2\pi}{\omega}\\\\[/tex]

[tex]\omega = \frac{2\pi}{T}[/tex]

where,

  • [tex]\omega[/tex] = angular speed = ?
  • T = time period = 2.2 x 10⁴ s

Therefore,

[tex]\omega=\frac{2\pi}{2.2\ x\ 10^4\ s}\\\\\omega=2.86\ x\ 10^{-4}\ rad/s[/tex]

Now, the tangential speed can be given by the following formula:

[tex]v=r\omega[/tex]

where,

  • r = radius = 1.8 x 10⁸ m
  • v = tangential speed = ?

Therefore,

[tex]v=(1.8\ x\ 10^8\ m)(2.86\ x\ 10^{-4}\ rad/s)[/tex]

v = 5.1 x 10⁴ m/s

Learn more about tangential speed here:

https://brainly.com/question/17025846

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