What is the acceleration due to gravity at a height h (= radius of the earth from the surface of the earth?

Answer the following question in detail.

Show that acceleration due to gravity at height h above the Earth’s surface is `"g"_"h" = "g"("R"/"R + h")^2`

Answer the following question in detail.

Discuss the variation of acceleration due to gravity with altitude.

1. Let, R = radius of the Earth, M = mass of the Earth.

   g = acceleration due to gravity at the surface of the Earth.

2. Consider a body of mass m on the surface of the Earth. The acceleration due to gravity on the Earth’s surface is given by,`"g" = "GM"/"R"^2`     ....(1)
   

3. The body is taken at height h above the surface of the Earth as shown in the figure. The acceleration due to gravity now changes to,
   `"g"_"h" = "GM"/("R + h")^2`     .....(2)
4. Dividing equation (2) by equation (1), we get,`"g"_"h"/"g" = ("GM"/("R + h")^2)/("GM"/"R"^2)``therefore "g"_"h"/"g" = "R"^2/("R + h")^2``therefore "g"_"h" = "gR"^2/("R + h")^2`We can rewrite,`therefore "g"_"h" = "gR"^2/("R"^2(1 + "h"/"R")^2)`

   `therefore "g"_"h" = "g"(1 + "h"/"R")^-2`

5. For small altitude h, i.e., for `"h"/"R"` << 1, by neglecting higher power terms of `"h"/"R"`, `"g"_"h" = "g"(1 - "2h"/"R")`

This expression can be used to calculate the value of g at height h above the surface of the Earth as long as h << R.

Concept: Variation in the Acceleration Due to Gravity with Altitude, Depth, Latitude and Shape

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