Question

Assume that the moon is hit by about 25 million micrometeorite impacts each day, and that these impacts strike randomly around the moon's surface. also assume that after a footprint is hit by 20 micrometeorites it is erased. (a) roughly (order of magnitude) how long do we expect it would take, in years, for one of the footprints left by the apollo astronauts to be completely erased?

Answer 1

The time required for the footprint left by the Apollo astronaut to be erased is $\fbox{\begin\\1010\,{\text{million}}\,{\text{yr}}\end{minispace}}$.

Further explanation:

Some astronauts went to the moon to know about the moon. They walk on the surface of moon that’s why their footprints left there. It took time to erase the footprints.

Given:

The number of micrometeorites hitting the moon in one day is $25 \times {10^6}$.

The area of footprint is $0.03\,{{\text{m}}^{\text{2}}}$.

The surface area of moon is $3.79 \times {10^{19}}\,{{\text{m}}^{\text{2}}}$.

Concept used:

When footprints hit the surface of moon it got impacted.

The expression for the number of micrometeorites hitting the surface area of footprints in one day is given as.

$\fbox{\begin\\n = \dfrac{N}{{{A_{{\text{moon}}}}}}{A_{{\text{footprint}}}}\end{minispace}}$                                   …… (1)

Here, $N$ is the number of micrometeorites hitting the moon in one day, ${A_{{\text{moon}}}}$ is the surface area of moon and ${A_{{\text{footprint}}}}$ is the surface area of footprint.

The expression for the time to receive impacts on footprints.

$\fbox{\begin\\T = \dfrac{{20}}{n}\end{minispace}}$                                                                                 …… (2)

Substitute $25 \times {10^6}$ for $N$, $0.03\, {{\text{m}}^{\text{2}}}$ for ${A_{{\text{footprint}}}}$ and in equation (1).

$\begin{aligned}n &= \frac{{\left( {25 \times {{10}^6}} \right)}}{{\left( {3.79 \times {{10}^{19}}\,{{\text{m}}^{\text{2}}}}\right)}}\left({0.03\,{{\text{m}}^{\text{2}}}\right)\\&=1.979\times{10^{-8}}\\\end{aligned}$

This means that the footprints receive $1.979 \times {10^{ - 8}}$ number of micrometeorites or impacts per day.

$20$ Impacts are required to erase the footprint in one day.

The expression for the time required to erase the footprints is given in equation (2).

Substitute $1.979 \times {10^{ - 8}}$for $n$ in equation (2).

$\begin{aligned}T &= \frac{{20}}{{\left( {1.979 \times {{10}^{-8}}}\right)}}\\&=1.010\times{10^9}\,{\text{days}}\\\end{aligned}$

Thus, it can be given in million years as $\fbox{1010\,{\text{million}}\,{\text{yr}}}$.

Learn more:

1.  Motion of ball under gravity brainly.com/question/10934170.

2.  Examples of wind and solar energy brainly.com/question/1062501.

3. Conservation of momentum brainly.com/question/9575487.

Answer Details:

Grade: College

Subject: Physics

Chapter: Astronomy  

Keywords:

Moon, meteorites, footprints, impact, surface area, 20 impacts, micrometeorites, 25 million  micrometeorites, 1010 million year,1.010*10^9 days .

Answer 2

Use the following formula to find the time meteorites take for a footprint to be erased:

T = [(Impact needed of a footprint)(4πr^2)]/[(impact rate)(Area of a footprint)].

 

Use the following formula to find the area of a foot:

Area of a foot = length of foot x width of foot

 

Substitute the given values in the above equation to get area of a foot:

Area of a foot = 25 cm x 10 cm = 250 cm^2

 

Convert the radius of moon in kilometers to centimeters.

R = 1738 km = 1738 km (10^5 cm / km) = 1.738 x 10^8 cm

 

The time meteorites take for a footprint to be erased:

T = (20)(4 π (1.738 x 10^8)^2) / (25 million / day) (250 cm^2) = 758.78 x 10^16 / 6250 x 10^6 /day = 1.21 x 10^9 days

 

Therefore, the time meteorites take for a footprint to be erased is 1.21 x 10^9 days