To find the tangential speed of an object orbiting Earth given a radius of [tex]\(1.8 \times 10^8 \)[/tex] meters and a period of [tex]\(2 \times 10^4 \)[/tex] seconds, we can use the formula for tangential speed in circular motion:
[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,
- [tex]\(\pi \)[/tex] is approximately 3.14159.
Let's break down the problem and solve it step by step.
1. Identify the given values:
- Radius [tex]\( r = 1.8 \times 10^8 \)[/tex] meters
- Period [tex]\( T = 2 \times 10^4 \)[/tex] seconds
2. Substitute the given values into the formula:
[tex]\[ v = \frac{2 \pi \cdot (1.8 \times 10^8)}{2 \times 10^4} \][/tex]
3. Simplify the expression inside the numerator:
[tex]\[ 2 \pi \cdot (1.8 \times 10^8) \][/tex]
This gives:
[tex]\[ 2 \cdot 3.14159 \cdot 1.8 \times 10^8 \][/tex]
[tex]\[ = 11.30973 \times 10^8 \][/tex]
4. Now, divide by the period:
[tex]\[ v = \frac{11.30973 \times 10^8}{2 \times 10^4} \][/tex]
5. Perform the division:
[tex]\[ v = \frac{11.30973 \times 10^8}{2 \times 10^4} \][/tex]
[tex]\[ = \frac{11.30973}{2} \times 10^{8-4} \][/tex]
[tex]\[ = 5.654865 \times 10^4 \][/tex]
Given the choices:
- [tex]\(7.7 \times 10^{-4} \text{ m/s}\)[/tex]
- [tex]\(5.1 \times 10^4 \text{ m/s}\)[/tex]
- [tex]\(7.7 \times 10^4 \text{ m/s}\)[/tex]
- [tex]\(5.1 \times 10^5 \text{ m/s}\)[/tex]
We see that the closest answer to our calculation of [tex]\(5.654865 \times 10^4 \text{ m/s}\)[/tex] is [tex]\(5.1 \times 10^4 \text{ m/s}\)[/tex].
Thus, the correct choice is:
[tex]\[ \boxed{5.1 \times 10^4 \text{ m/s}} \][/tex]
