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What is the approximate tangential speed of an object orbiting Earth with a radius of [tex][tex]$1.8 \times 10^8 \, \text{m}$[/tex][/tex] and a period of [tex]2.2 \times 10^4 \, \text{s}$[/tex]?

A. [tex]7.7 \times 10^{-4} \, \text{m/s}[/tex]
B. [tex]5.1 \times 10^4 \, \text{m/s}[/tex]
C. [tex]7.7 \times 10^4 \, \text{m/s}[/tex]
D. [tex]5.1 \times 10^5 \, \text{m/s}[/tex]



Answer :

GinnyAnswer
To find the tangential speed of an object orbiting Earth, we use the formula for tangential speed:

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

where:
- [tex]\( v \)[/tex] is the tangential speed
- [tex]\( r \)[/tex] is the radius of the orbit
- [tex]\( T \)[/tex] is the period of the orbit

Given:
[tex]\[ r = 1.8 \times 10^8 \text{ meters} \][/tex]
[tex]\[ T = 2.2 \times 10^4 \text{ seconds} \][/tex]

1. First, plug the values into the formula:

[tex]\[ v = \frac{2 \pi \times 1.8 \times 10^8 \text{ m}}{2.2 \times 10^4 \text{ s}} \][/tex]

2. Calculate the numerator [tex]\( 2 \pi \times 1.8 \times 10^8 \)[/tex]:

[tex]\[ 2 \pi \times 1.8 \times 10^8 \approx 11.309733552923255 \times 10^8 \text{ m} \][/tex]

3. Next, divide this result by the period [tex]\( 2.2 \times 10^4 \)[/tex]:

[tex]\[ \frac{11.309733552923255 \times 10^8 \text{ m}}{2.2 \times 10^4 \text{ s}} \approx \frac{11.309733552923255 \times 10^8}{2.2 \times 10^4} \approx 51407.8797860148 \text{ m/s} \][/tex]

Hence, the approximate tangential speed of the object is:

[tex]\[ 51407.88 \text{ m/s} \][/tex]

Therefore, among the provided choices, the closest value is:

[tex]\[ 5.1 \times 10^4 \text{ m/s} \][/tex]

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