Suppose NASA wants a satellite to revolve around Earth 5 times a day. What should be the radius of its orbit if we neglect the presence of the Moon? $\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } , M _ { \text {earth } } = 5.97 \times \right.$ $10 ^ { 24 } \mathrm {~kg} \text { ) }$
A) $7.22 \times 10 ^ { 7 } \mathrm {~m}$
B) $2.11 \times 10 ^ { 7 } \mathrm {~m}$
C) $0.69 \times 10 ^ { 7 } \mathrm {~m}$
D) $1.44 \times 10 ^ { 7 } \mathrm {~m}$

Question

Quizplus · Accepted Answer

To find the radius of the orbit, use the formula for the orbital period $ T = 2\pi \sqrt{\frac{r^3}{GM}} $. Given that the satellite revolves 5 times a day, $ T = \frac{24 \times 3600}{5} $ seconds. Solving for $ r $ gives approximately $ 1.44 \times 10^7 $ meters.

Suppose NASA Wants a Satellite to Revolve Around Earth 5 (G=6.67×10−11 N⋅m2/kg2,Mearth =5.97×\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } , M _ { \text {earth } } = 5.97 \times \right.(G=6.67×10−11 N⋅m2/kg2,Mearth ​=5.97×

Suppose NASA Wants a Satellite to Revolve Around Earth 5 $\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } , M _ { \text {earth } } = 5.97 \times \right.$