Question

A moon lander is orbiting the moon at an altitude of 1000 km. By what percentage must it decrease its speed so as to just graze the moon’s surface one-half period later?

Short answer

The moon lander must decrease its speed by 11.8%

Step-by-step solution

Given information

A moon lander is orbiting the moon at an altitude of 1000 km .

Assume moon is spherical

Explanation

For the moon lander the energy and momentum both are conserved between the points while orbiting.

Lets first find the the original speed of the lander

$$\begin{gathered} v_o=\sqrt{\frac{{\mathrm{GM}}_{\mathrm{m}}}{{\mathrm{R}}_{\mathrm{m}}+\mathrm{h}}} \\ =1338\mathrm{m}/\mathrm{s} \end{gathered}$$

Next find the final speed and then get the % decrease.

From the conservation of angular momentum:

${\mathrm{mr}}_1{\mathrm{v}}_1=m{r}_2{\mathrm{v}}_2$

$\Rightarrow \frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}=\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{{\mathrm{R}}_{\mathrm{m}}+\mathrm{h}}{{\mathrm{R}}_{\mathrm{m}}}$

Now from energy conservation equation

$\frac{1}{2}\mathrm{m}{\left({\mathrm{v}}_1\right)}^2-\frac{{\mathrm{GM}}_{\mathrm{m}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{m}}+\mathrm{h}}=\frac{1}{2}\mathrm{m}{\left({\mathrm{v}}_2\right)}^2-\frac{{\mathrm{GM}}_{\mathrm{m}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{m}}}$

Now substitute the value of V2 , as ${\mathrm{v}}_2=\frac{\left({\mathrm{R}}_{\mathrm{m}}+\mathrm{h}/{\mathrm{R}}_{\mathrm{m}}\right)}{{\mathrm{v}}_1}$, we get

$\frac{1}{2}{\left(v_1\right)}^2-\frac{G{M}_m}{R_m+h}=\frac{1}{2}{\left(v_1\right)}^2{\left(\frac{R_m+h}{R_m}\right)}^2-\frac{G{M}_m}{R_m}$

\Substitute the values and solve for v₁

$$\begin{gathered} {\left({\mathrm{v}}_1\right)}^2=1.392\times {10}^6{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {v}_1=1180\mathrm{m}/\mathrm{s} \end{gathered}$$

Now find percent decrease

$$\begin{gathered} =\frac{(1338\mathrm{m}/\mathrm{s}-1180\mathrm{m}/\mathrm{s})}{1180\mathrm{m}/\mathrm{s}}\times 100\% \\ =11.8\% \end{gathered}$$

It need to reduce the speed by 11.8%