A satellite orbits the Earth once every 6.0 hours in a circle. What are the magnitude and direction of the acceleration of the satellite? $\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } , M _ { \text {earth } } = 5.97 \times \right.$ $\left. 10 ^ { 24 } \mathrm { kq } \right)$

Question

A satellite orbits the Earth once every  6.0  hours in a circle. What are the magnitude and direction of the acceleration of the satellite? $\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } , M _ { \text {earth } } = 5.97 \times \right.$ $\left. 10 ^ { 24 } \mathrm { kq } \right)$

Quizplus · Accepted Answer

The Answer of A satellite orbits the Earth once every  6.0  hours in a circle. What are the magnitude and direction of the acceleration of the satellite? $\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } , M _ { \text {earth } } = 5.97 \times \right.$ $\left. 10 ^ { 24 } \mathrm { kq } \right)$

A Satellite Orbits the Earth Once Every 6 (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×

A Satellite Orbits the Earth Once Every 6 $\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } , M _ { \text {earth } } = 5.97 \times \right.$