Chapter 13: Q. 70 (page 356)

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?

Expert verified

The moon lander must decrease its speed by 11.8%

Step by step solution

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

Assume moon is spherical

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

$v_{o} = \sqrt{\frac{{GM}_{m}}{R_{m} + h}} = 1338 m / s$

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

From the conservation of angular momentum:

${mr}_{1} v_{1} = m r_{2} v_{2}$

$\Rightarrow \frac{v_{2}}{v_{1}} = \frac{r_{1}}{r_{2}} = \frac{R_{m} + h}{R_{m}}$

Now from energy conservation equation

\frac{1}{2} m {(v_{1})}^{2} - \frac{{GM}_{m} m}{R_{m} + h} = \frac{1}{2} m {(v_{2})}^{2} - \frac{{GM}_{m} m}{R_{m}}

Now substitute the value of V2 , as $v_{2} = \frac{(R_{m} + h / R_{m})}{v_{1}}$ , we get

$\frac{1}{2} {(v_{1})}^{2} - \frac{G M_{m}}{R_{m} + h} = \frac{1}{2} {(v_{1})}^{2} {(\frac{R_{m} + h}{R_{m}})}^{2} - \frac{G M_{m}}{R_{m}}$

\Substitute the values and solve for v₁

${(v_{1})}^{2} = 1.392 \times 10^{6} m^{2} / s^{2} v_{1} = 1180 m / s$

Now find percent decrease

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

It need to reduce the speed by 11.8%

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