Physics · Ch 5 — Gravitation

(A) Variation in g with Altitude

5.6.1

(A) Variation in g with Altitude

Consider a body of mass m at the Earth's surface, where the acceleration due to gravity is $g=\dfrac{GM}{R^2}$ . When the same body is raised to height h above the surface (Fig. 5.6), its distance from the Earth's centre becomes $(R+h)$ , so the acceleration due to gravity there is $g_h=\dfrac{GM}{(R+h)^2}$ . Dividing this by the surface expression eliminates GM: $\dfrac{g_h}{g}=\dfrac{R^2}{(R+h)^2}$ , giving the exact result $g_h=\dfrac{gR^2}{(R+h)^2}.$ This shows g decreases steadily and continuously as altitude h increases -- there is no height, however large, at which g becomes exactly zero (it only approaches zero as h tends to infinity). …

Figure 5.6Fig. 5.6: Acceleration due to gravity at height h above the Earth's surface

What this figure shows. A circle representing the Earth, with centre O and radius R, is drawn with a point marked on its surface. A second point, representing an object at height h above the surface, is marked along the same radial line extended outward beyond the surface point, at a further distance h from it. The total distance from the Earth's centre O to this elevated point is labelled $R+h$ , and this is the effective distance r used in $g_h=GM/(R+h)^2$ -- the figure establishes that at height h, the relevant distance from the centre of mass is the Earth's radius PLUS the altitude …

Misc Ex.7Example 5.7: Height above the surface at which g decreases by 10%

Worked out. Using the small-altitude approximation $g_h=g(1-2h/R)$ with $g_h=0.9g$ (a 10% decrease) and R = 6400 km: $0.9=1-2h/R\Rightarrow2h/R=0.1\Rightarrow h=R/20=6400/20=320$ km. The example is a direct application of the linearised altitude-variation formula, valid since 320 km is indeed small compared to Earth's 6400 km radius. …

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