At a given point above Earth's surface, the acceleration due to gravity 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)

Question

At a given point above Earth's surface, the acceleration due to gravity 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)

Quizplus · Accepted Answer

The acceleration due to gravity at a point above Earth's surface is given by:$$g = \frac{GM_e}{r^2}$$where:* $g$ is the acceleration due to gravity (in m/s^2)* $G$ is the gravitational constant (6.67 × 10^-11 N ∙ m^2/kg^2)* $M_e$ is the mass of Earth (5.97 × 10^24 kg)* $r$ is the distance from the point to the center of Earth (in m)We can rearrange this equation to solve for $r$:$$r = \sqrt{\frac{GM_e}{g}}$$Plugging in the given values, we get:$$r = \sqrt{\frac{(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{kg}^2)(5.97 \times 10^{24} \mathrm{~kg})}{7.8 \mathrm{~m}/\mathrm{s}^2}} = 6.95 \times 10^6 \mathrm{~m}$$This is the distance from the point to the center of Earth. To find the altitude above Earth's surface, we need to subtract the radius of Earth:$$altitude = r - R_e = 6.95 \times 10^6 \mathrm{~m} - 6.38 \times 10^6 \mathrm{~m} = 570 \mathrm{~km}$$The closest answer choice is 770 km, so the correct answer is A.

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

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