the orbital period of a satellite is 2×10^6 s and its total radius is 2.5×10^12 m. the tangential speed of…

the orbital period of a satellite is 2×10^6 s and its total radius is 2.5×10^12 m. the tangential speed of the satellite is m/s. (round to the nearest whole number, do not add any punctuation.)

the orbital period of a satellite is 2×10^6 s and its total radius is 2.5×10^12 m. the tangential speed of the satellite is m/s. (round to the nearest whole number, do not add any punctuation.)

Answer

Answer:

7854

Explanation:

Step1: Calculate the circumference

$C = 2\pi r$ $C=2\times\pi\times2.5\times 10^{12}$

Step2: Find the tangential speed

$v=\frac{C}{T}$ $v=\frac{2\times\pi\times2.5\times 10^{12}}{2\times 10^{6}}$ $v=\pi\times2.5\times 10^{6}$ $v\approx 7854$