To determine the tangential speed of an object orbiting Earth, we can 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.
Let's plug in the values provided in the problem:
- Radius ([tex]\(r\)[/tex]) = [tex]\( 1.8 \times 10^8 \)[/tex] meters,
- Period ([tex]\(T\)[/tex]) = [tex]\( 2.2 \times 10^4 \)[/tex] seconds.
Now, substituting these numbers into the formula:
[tex]\[ v = \frac{2 \pi \times 1.8 \times 10^8 \, \text{m}}{2.2 \times 10^4 \, \text{s}} \][/tex]
We need to calculate the numerator first:
[tex]\[ 2 \pi \times 1.8 \times 10^8 \][/tex]
Multiplying [tex]\(2 \pi \approx 6.2832\)[/tex] by [tex]\(1.8 \times 10^8\)[/tex]:
[tex]\[ 2 \pi \times 1.8 \times 10^8 \approx 6.2832 \times 1.8 \times 10^8 \][/tex]
[tex]\[ \approx 11.30976 \times 10^8 \][/tex]
Next, we divide this value by the period:
[tex]\[ v = \frac{11.30976 \times 10^8 \, \text{m}}{2.2 \times 10^4 \, \text{s}} \][/tex]
Performing the division:
[tex]\[ v \approx \frac{11.30976 \times 10^8}{2.2 \times 10^4} \][/tex]
[tex]\[ v \approx 5.140787878 \times 10^4 \, \text{m/s} \][/tex]
Rounding this value to two significant figures:
[tex]\[ v \approx 5.1 \times 10^4 \, \text{m/s} \][/tex]
So, the approximate tangential speed of the object is:
[tex]\[ \boxed{5.1 \times 10^4 \, \text{m/s}} \][/tex]
