Chapter 13.1, Problem 13.36P

A meteor starts from rest at a very great distance from the earth. Knowing that the radius of the earth is 6370 km and neglecting all forces except the gravitational attraction of the earth, determine the speed of the meteor (a) when it enters the ionosphere at an altitude of 1000 km. (b) when it enters the stratosphere at an altitude of 50 km. (c) when it strikes the earth’s surface.

To determine

(a)

Speed of meteor.

Answer to Problem 13.36P

Speed of meteor is, v=10.39 km/s.

Explanation of Solution

Given information:

Initially start from rest.

Radius of earth (R) = 6370 km.

When it enters at ionosphere at an altitude of 1000km.

Calculation:

From work energy principle for a meteor,

$-Fdr=mvdv$

Where, F is gravitational force and given as

$F=\frac{gm{R}^2}{r^2}$ Where R radius of earth, r distance travel by the meteor and m mass of the meteor.

So, work energy equation

$\begin{array}{l}-\frac{gm{R}^2}{r^2}dr=mvdv\\ vdv=-\frac{g{R}^2}{r^2}dr\end{array}$

Integrate the equation from o to v and 0 to x.

$\begin{array}{l}{f}_o{}^{v}vdv=-f_0{}^x\frac{g{R}^2}{r^2}\\ \frac{v^2}{2}=\frac{g{R}^2}{x}\\ v=\sqrt{\frac{2g{R}^2}{x}}\end{array}$

This is the equation for velocity of the meteor.

For x=6370km+1000 km=7370km

$\begin{array}{l}v=\sqrt{\frac{2g{R}^2}{x}}\\ v=\sqrt{\frac{2\times 9.81\times {\left(6370\times 1000m\right)}^2}{\left(7370\times 1000m\right)}}\\ v=\sqrt{\frac{19.62\times {\left(6370\times 1000m\right)}^2}{\left(7370\times 1000m\right)}}\\ v=\frac{4.43\times 6370\times 1000}{2714.774}\\ v=103934m/s\\ v=10.39km/s\end{array}$

To determine

(b)

Speed of meteor

Answer to Problem 13.36P

Speed of meteor is, v = 11.137 km/s.

Explanation of Solution

Given information:

Initially start from rest.

Radius of earth (R) = 6370 km.

When it enters at ionosphere at an altitude of 50km.

Calculation:

From work energy principle for a meteor

$-Fdr=mvdv$

Where, F is gravitational force and given as

$F=\frac{gm{R}^2}{r^2}$

Where, R radius of earth, r distance travel by the meteor and m mass of the meteor.

So, work energy equation

$\begin{array}{l}-\frac{gm{R}^2}{r^2}dr=mvdv\\ vdv=-\frac{g{R}^2}{r^2}dr\end{array}$

Integrate the equation from o to v and 0 to x.

$\begin{array}{l}{f}_o{}^{v}vdv=-f_0{}^x\frac{g{R}^2}{r^2}\\ \frac{v^2}{2}=\frac{g{R}^2}{x}\\ v=\sqrt{\frac{2g{R}^2}{x}}\end{array}$

This is the equation for velocity of the meteor.

For x=6370km+50 km=6420 km

$\begin{array}{l}v=\sqrt{\frac{2g{R}^2}{x}}\\ v=\sqrt{\frac{2\times 9.81\times {\left(6370\times 1000m\right)}^2}{\left(6420\times 1000m\right)}}\\ v=\sqrt{\frac{19.62\times {\left(6370\times 1000m\right)}^2}{\left(6420\times 1000m\right)}}\\ v=\frac{4.43\times 6370\times 1000}{2533.77}\\ v=11137.2m/s\\ v=11.137km/s\end{array}$

To determine

(c)

Speed of meteor.

Answer to Problem 13.36P

Speed of meteor when it strike to the earth is, v=11.18 km/s.

Explanation of Solution

Given information:

Initially start from rest.

Radius of earth (R) = 6370 km.

When it strikes to the earth.

Calculation:

From work energy principle for a meteor

$-Fdr=mvdv$

Where, F is gravitational force and given as

$F=\frac{gm{R}^2}{r^2}$

Where, R radius of earth, r distance travel by the meteor and m mass of the meteor.

So, work energy equation

$\begin{array}{l}-\frac{gm{R}^2}{r^2}dr=mvdv\\ vdv=-\frac{g{R}^2}{r^2}dr\end{array}$

Integrate the equation from o to v and 0 to x.

$\begin{array}{l}{f}_o{}^{v}vdv=-f_0{}^x\frac{g{R}^2}{r^2}\\ \frac{v^2}{2}=\frac{g{R}^2}{x}\\ v=\sqrt{\frac{2g{R}^2}{x}}\end{array}$

This is the equation for velocity of the meteor.

For x=6370km+0 km=6370km,

$\begin{array}{l}v=\sqrt{\frac{2g{R}^2}{x}}\\ v=\sqrt{\frac{2\times 9.81\times {\left(6370\times 1000m\right)}^2}{\left(6370\times 1000m\right)}}\\ v=\sqrt{\frac{19.62\times {\left(6370\times 1000m\right)}^2}{\left(6370\times 1000m\right)}}\\ v=\frac{4.43\times 6370\times 1000}{2523.88}\\ v=11180.84m/s\\ v=11.18km/s\end{array}$