Certainly! Let's solve for the approximate tangential speed of an object orbiting Earth given the radius and the period of the orbit.
### Step-by-Step Solution:
1. Given:
- Radius of orbit ([tex]\(r\)[/tex]) = [tex]\(1.8 \times 10^8\)[/tex] meters
- Period of orbit ([tex]\(T\)[/tex]) = [tex]\(2.2 \times 10^4\)[/tex] seconds
2. Calculate the Circumference of the Orbit:
The formula for the circumference [tex]\(C\)[/tex] of a circle (which, in this case, is the orbit) is:
[tex]\[
C = 2 \pi r
\][/tex]
Plugging in the given radius:
[tex]\[
C = 2 \pi \times 1.8 \times 10^8 \text{ meters}
\][/tex]
This results in:
[tex]\[
C \approx 1,130,973,355.2923255 \text{ meters}
\][/tex]
3. Calculate the Tangential Speed ([tex]\(v\)[/tex]):
The tangential speed of an object in circular motion is given by the formula:
[tex]\[
v = \frac{C}{T}
\][/tex]
where:
- [tex]\(C\)[/tex] is the circumference of the orbit
- [tex]\(T\)[/tex] is the period of the orbit
Plugging in the values for circumference and period:
[tex]\[
v = \frac{1,130,973,355.2923255 \text{ meters}}{2.2 \times 10^4 \text{ seconds}}
\][/tex]
Simplifying this:
[tex]\[
v \approx 51407.8797860148 \text{ meters per second (m/s)}
\][/tex]
4. Compare the Calculated Tangential Speed to Given Options:
The given multiple-choice options are:
- [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]
Our calculated tangential speed ([tex]\(v \approx 51407.8797860148 \text{ m/s}\)[/tex]) can be approximated as [tex]\( 5.1 \times 10^4 \text{ m/s}\)[/tex].
Thus, the approximate tangential speed of the object is:
[tex]\[
\boxed{5.1 \times 10^4 \text{ m/s}}
\][/tex]
