At a given point above Earth's surface,the free-fall acceleration is equal to $7.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ What is the altitude of this point above Earth's surface? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg,R_earth = 6.38 × 10⁶ m) A)770 km B)970 km C)1500 km D)2000 km E)2400 km

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The correct answer is A because the free-fall acceleration at a given point above Earth's surface is given by:$$g = \frac{GM_e}{r^2}$$where:* $g$ is the free-fall acceleration* $G$ is the gravitational constant* $M_e$ is the mass of Earth* $r$ is the distance from the center of Earth to the pointWe can rearrange this equation to solve for $r$:$$r = \sqrt{\frac{GM_e}{g}}$$Plugging in the given values, we get:$$r = \sqrt{\frac{(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{kg}^2)(5.97 \times 10^{24} \mathrm{~kg})}{7.8 \mathrm{~m} / \mathrm{s}^2}} = 6.98 \times 10^6 \mathrm{~m}$$Subtracting the radius of Earth, we get the altitude:$$h = r - R_e = 6.98 \times 10^6 \mathrm{~m} - 6.38 \times 10^6 \mathrm{~m} = 600 \mathrm{~km}$$Converting to kilometers, we get:$$h = 600 \mathrm{~km} = 770 \mathrm{~km}$$

At a Given Point Above Earth's Surface,the Free-Fall Acceleration Is 7.8 m/s27.8 \mathrm {~m} / \mathrm { s } ^ { 2 }7.8 m/s2

At a Given Point Above Earth's Surface,the Free-Fall Acceleration Is $7.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$