Advertisement Remove all ads

Answer in Brief

**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.

Advertisement Remove all ads

#### Solution

- Let,

R = radius of the Earth,

M = mass of the Earth.

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

Is there an error in this question or solution?

Advertisement Remove all ads

Advertisement Remove all ads