We are given that an object leaves the earth at a speed of 5054 m/s. We want to determine how far from the center of the earth the object travels. To do that we will do an energy balance. We will have into account that the total energy at the surface of the earth is the kinetic energy of the object plus the gravitational potential energy, and when the object stops the energy is equal to the gravitational potential energy:
[tex]\begin{gathered} E_1=E_k+E_{g1} \\ E_2=E_{g1} \end{gathered}[/tex]
Where:
[tex]\begin{gathered} E_1=\text{ energy at the surface of the earth} \\ E_k=\text{ kinetic energy} \\ E_g=\text{ }gravitational\text{ potential energy} \\ E_2=\text{ energy when the object stops} \end{gathered}[/tex]
The kinetic energy is given by:
[tex]E_k=\frac{1}{2}mv^2[/tex]
Where:
[tex]\begin{gathered} m=\text{ mass} \\ v=\text{ velocity} \end{gathered}[/tex]
The gravitational potential energy is given by:
[tex]E_g=-G\frac{Mm}{R}[/tex]
Where:
[tex]\begin{gathered} G=\text{ gravitational contant} \\ M=\text{ mass of th earth} \\ m=\text{ mass of the object} \\ R=\text{ distance to the center of the earth at the given point} \end{gathered}[/tex]
Substituting the values we get:
[tex]\begin{gathered} E_1=\frac{1}{2}mv^2-G\frac{Mm}{R_1} \\ \\ E_2=-G\frac{Mm}{R_2} \end{gathered}[/tex]
In this case, R1 is the radius of the earth since the object leaves from the surface of the earth and R2 the distance to the center of the earth from the object stops as shown in the diagram above.
Now we can set them equal due to conservation of energy:
[tex]\frac{1}{2}mv^2-G\frac{Mm}{R_1}=-G\frac{Mm}{R_2}[/tex]
Now we may cancel out the "m":
[tex]\frac{1}{2}v^2-G\frac{M}{R_1}=-G\frac{M}{R_2}[/tex]
We wish to determine the distance to the center of the earth when the object stops, therefore, we solve for "R2". We do this first by multiplying both sides by R2:
[tex]R_2(\frac{1}{2}v^2-G\frac{M}{R_1})=-GM[/tex]
Now we divide both sides by the factor multiplying R2:
[tex]R_2=-\frac{GM}{(\frac{1}{2}v^2-G\frac{M}{R_1})}[/tex]
Now we substitute the known values:
[tex]R_2=-\frac{(6.67428\times10^{-11}\frac{Nm^2}{\operatorname{kg}^2})(5.9742\times10^{24}\operatorname{kg})}{(\frac{1}{2}(5054\frac{m}{s})^2-((6.67428\times10^{-11}\frac{Nm^2}{\operatorname{kg}^2})\frac{5.9742\times10^{24}\operatorname{kg}}{(6.3781\times10^6_{}m)})}[/tex]
Solving the operations we get:
[tex]R_2=8.06\times10^6m[/tex]
Therefore, the distance from the earth is 8.06x10^6 meters.
