Answer to Question #96628 in Classical Mechanics for matilde ferretti

Question

At a distance H above the surface of a planet, the true weight of a remote probe is eight percent less than its true weight on the surface. The radius of the planet is R. Find the ratio H/R.
H
R
=
1.24

Answer 1

Due to the Newton's law of gravity the gravitational acceleration is (we assume $r \ge R$ , where $R$ - radius of the planet, also we assume that the planet has a spherical form)

$$\left| {\vec g} \right| = {{GM} \over {{r^2}}}$$

where $G$ - gravitational constant, $M$ - mass of the planet and $r$ - distance. Then at the surface of the planet

$$g = {{GM} \over {{R^2}}}$$

and at the distance H above the surface

$$g' = {{GM} \over {{{(R + H)}^2}}}$$

According that the weight is $\vec P = m\vec g$ and the mass of the body $m$ does not change, we know from the conditions of the problem that $g' = kg = 0.92g$ .

Then we have

$${{GM} \over {{{(R + H)}^2}}} = k{{GM} \over {{R^2}}}$$ $${(R + H)^2} = {1 \over k}{R^2}$$ $${(1 + {H \over R})^2} = {1 \over k}$$

and (we reject 'unphysical' root)

$${H \over R} = {1 \over {\sqrt k }} - 1$$

In our case we get

$${H \over R} = {1 \over {\sqrt {0.92} }} - 1 \approx 0.0426$$