Answer:
circumference of the satellite orbit = 4.13 × 10⁷ m
Step-by-step explanation:
Given that:
the time period T = 88.5 min = 88.5 × 60 = 5310 sec
The mass of the earth
= 5.98 × 10²⁴ kg
if the radius of orbit is r,
Then,
![(V^2)/(r) = (GM_e)/(r^2)](https://img.qammunity.org/2021/formulas/physics/college/9yc7lr9t810o5fi3408vkb45i7k0b8bwht.png)
![{V^2} = (GM_e r)/(r^2)](https://img.qammunity.org/2021/formulas/physics/college/4j2p2yp7m381hx9w4132gtz5j0hn50grg9.png)
![{V^2} = (GM_e )/(r)](https://img.qammunity.org/2021/formulas/physics/college/qffbe8kjtn4cm4fb4k4itmgfr1n5gygowf.png)
![{V} =\sqrt{ (GM_e )/(r)}](https://img.qammunity.org/2021/formulas/physics/college/2ws6yepwp2u0rcuzxtitb2tbchrmm5cxog.png)
Similarly :
![T = \sqrt{\frac{ 2 \pi r} {V} }](https://img.qammunity.org/2021/formulas/physics/college/pascw1oe0441g6siu0in2ttp28dx54kpbg.png)
where;
![{V} =\sqrt{ (GM_e )/(r)}](https://img.qammunity.org/2021/formulas/physics/college/2ws6yepwp2u0rcuzxtitb2tbchrmm5cxog.png)
Then:
![T = {\frac{ 2 \pi r^(3/2)} {\sqrt{ {GM_e }} }](https://img.qammunity.org/2021/formulas/physics/college/d0voxrsp5eeqrunrbhzgvuekdrayet8yeb.png)
![5310= {\frac{ 2 \pi r^(3/2)} {\sqrt{ {6.674* 10^(-11) * 5.98 * 10^(24) }} }](https://img.qammunity.org/2021/formulas/physics/college/qmamgbfmw18ecoe3gsxj8ck6m6gk0lhmhn.png)
![5310= {\frac{ 2 \pi r^(3/2)} {\sqrt{ 3.991052 * 10^(14) }}](https://img.qammunity.org/2021/formulas/physics/college/d12tlaa23t9ygp4sfqd9as7kxim8e1kckt.png)
![5310= {\frac{ 2 \pi r^(3/2)} {19977617.48}](https://img.qammunity.org/2021/formulas/physics/college/o7vkxi92rs2dzmgwf3eri74xwp625w9hxq.png)
![5310 * 19977617.48= 2 \pi r^(3/2)}](https://img.qammunity.org/2021/formulas/physics/college/7x77v37aypnz3yb5daa1mzhuqwp3jwi34w.png)
![1.06081149 * 10^(11)= 2 \pi r^(3/2)}](https://img.qammunity.org/2021/formulas/physics/college/fvmuhd3279596wat79lui7ele9keg7rc5k.png)
![(1.06081149 * 10^(11))/(2 \pi)= r^(3/2)}](https://img.qammunity.org/2021/formulas/physics/college/g9hqfe97ejbbx3t3gmv5iqu0br2c29o9s3.png)
![r^(3/2)} = (1.06081149 * 10^(11))/(2 \pi)](https://img.qammunity.org/2021/formulas/physics/college/nx8kucl6hkouwrzd9vkp5pn7t7csiyd6de.png)
![r^(3/2)} = 1.68833392 * 10^(10)](https://img.qammunity.org/2021/formulas/physics/college/ozqqarl6m85ph43h6iwlcelntg4wf7mbds.png)
![r= (1.68833392 * 10^(10))^(2/3)}](https://img.qammunity.org/2021/formulas/physics/college/yyxitydg9disjv4xbm1jg18uwbhwbquu00.png)
![r= 2565.38^2](https://img.qammunity.org/2021/formulas/physics/college/t3o7a3w94cmpaf3107qwwqfhix1itqscag.png)
r = 6579225 m
The circumference of the satellites orbit can now be determined by using the formula:
circumference = 2π r
circumference = 2π × 6579225 m
circumference = 41338489.85 m
circumference of the satellite orbit = 4.13 × 10⁷ m