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At a Given Point Above Earth's Surface, the Acceleration Due 7.8 m/s27.8 \mathrm {~m} / \mathrm { s } ^ { 2 }

question 20

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At a given point above Earth's surface, the acceleration due to gravity is equal to 7.8 m/s27.8 \mathrm {~m} / \mathrm { s } ^ { 2 }
the altitude of this point above Earth's surface? (G=6.67×1011 Nm2/kg2,Mearth =5.97×1024\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } , M _ { \text {earth } } = 5.97 \times 10 ^ { 24 } \right. kg,Rearth =6.38×106 m )  \mathrm { kg } , R _ { \text {earth } } = 6.38 \times 10 ^ { 6 } \mathrm {~m} \text { ) }


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