Answer to Question #86969 in Mechanics | Relativity for Maria

Question

You are explaining to friends why an astronaut feels weightless orbiting in the space shuttle, and they respond that they thought gravity was just a lot weaker up there.
Convince them that it isn't so by calculating how much weaker (in %) gravity is 440 km above the Earth's surface.
Express your answer using three significant figures.

Answer 1

We can find the acceleration of gravity at any height above the Earth’s surface from the formula:

$$g_E = G \dfrac{M_E}{(R_E + h)^2},$$

here, $G$ is the gravitational constant, $M_E=5.98 \cdot 10^{24} kg$ is the mass of the Earth, $R_E=6.38 \cdot 10^6 m$ is the radius of the Earth and $h$ is the height above the Earth’s surface.

Let’s calculate the acceleration of gravity at $440 km$ above the Earth’s surface:

$$g_{440 km} = 6.67 \cdot 10^{-11} \dfrac{N \cdot m^2}{kg^2} \cdot \dfrac{5.98 \cdot 10^{24} kg}{(6.38 \cdot 10^6 m + 4.4 \cdot 10^5 m)^2} = 8.57 \dfrac{m}{s^2}.$$

Let’s compare (in %) the acceleration of gravity at $440 km$ above the Earth’s surface to the acceleration of gravity at the Earth’s surface:

$$\dfrac{g_{440 km}}{g_{Earth's surface}} = \dfrac{8.57 \dfrac{m}{s^2}}{9.8 \dfrac{m}{s^2}} \cdot 100 \% = 87.4 \%.$$

Answer:

Therefore, the gravity is 87.4% as strong as at the Earth’s surface.