Question

a body is taken to a height equal to the radius of the Earth. if acceleration due to to the gravity on the surface of the earth is is 9.8 metre per second square .what will be the the acceleration due to to the height where the body is taken?​

Answer 1

Given:-

• Radius of earth = 6400km

• Body is taken to height equal to the radius of the earth.

• Acceleration due to Gravity on the surface of the earth is 9.8m/s

To Find:-

• The Acceleration due to height where the body is taken.

Formulae used:-

• $\rm{g_{earth}}$= $\rm{G\times{\dfrac{M}{r²}}}$

• $\rm{g_{height} = G\times{\dfrac{M}{(r+h)^2}}}$

Now,

As we know the equation to find the Acceleration due to Gravity on the surface of the earth.

→ $\rm{g_{earth}}$= $\rm{G\times{\dfrac{M}{r^2}}}$.........1

According to Question, Height = Radius of the

earth

Therefore,

→ $\rm{g_{height} = G\times{\dfrac{M}{(r+h)^2}}}$.........2

Now, Dividing eq.1 and eq.2

→ $\rm{\dfrac{g_{height}{g_{earth} = \dfrac{G\times{\dfrac{M}{(r+h)^{2}}}{G\times{\dfrac{M}{r^{2}}}}$

→ $\rm{\dfrac{g_height}{g_earth} = \dfrac{r^2}{(r+h)^2}}$

→ $\rm{g_{height} = g_{earth}\times{\dfrac{r^2}{(r+h)^2}}}$

→ $\rm{g_{height} = 9.8\times{\dfrac{(6400)^2}{(6400 + 6400)^2}}}$

→ $\rm{ g_{height} = 9.8\times{\dfrac{4096000}{163840000}}}$

→ $\rm{g_{height} = 9.8\times{0.025}}$

→ $\rm{g_{height} = 0.245m/s^2}$.

Hence, the Acceleration due to Gravity will be 0.245m/s²..