Question

A rocket is launched from the surface of the earth with a speed of 9.0x103 m/s. What is the maximum altitude reached by the rocket? (MEarth=5.98x1024 kg, REarth=6.37x106 m)

Answer 1

From the Law of conservation of energy, we know that the sum of the kinetic and potential energy of the rocket is the same at the surface of the Earth and at the maximum altitude. Nevertheless, the kinetic energy of the rocket when it is at the maximum altitude is 0:

`\begin{gathered} K_1+U_1=K_2+U_2 \\ K_2=0 \\ \Rightarrow K_1+U_1=U_2 \end{gathered}`

The kinetic energy is given by:

`K=(1)/(2)mv^2`

On the other hand, the gravitational potential energy for big changes in altitude (comparable to the radius of the Earth) is given by the expression:

`U=-(GMm)/(r)`

Where M is the mass of the Earth, m is the mass of the rocket, r is the distance from the center of the Earth to the rocket and G is the gravitational constant:

At the beggining of the movement, the value of r corresponds to the radius of the Earth:

`U_1=-(GMm)/(R_E)`

At the end of the movement, the value of r corresponds to the radius of the Earth plus the maximum altitude h:

`U_2=-\frac{GMm}{R_E+h_{}}`

Substitute the expressions for U_1, K_1 and U_2 and simplify the equation by eliminating the factor m:

`\begin{gathered} (1)/(2)mv^2-(GMm)/(R_E)=-(GMm)/(R_E+h) \\ \Rightarrow(1)/(2)v^2-(GM)/(R_E)=-(GM)/(R_E+h) \end{gathered}`

Isolate the term GM/(R_E+h):

`\Rightarrow(GM)/(R_E+h)=\frac{GM}{R_E_{}}-(1)/(2)v^2`

Divide both sides by the factor GM:

`\Rightarrow(1)/(R_E+h)=(1)/(R_E)-(v^2)/(2GM)`

Take the reciprocal to both sides of the equation:

`\Rightarrow R_E+h=(1)/((1)/(R_E)-(v^2)/(2GM))`

Isolate h:

`h=(1)/((1)/(R_E)-(v^2)/(2GM))-R_E`

Substitute the values of each variable: R_E=6.37x10^6m, M=5.98x10^24kg, G=6.67x10^-11 N*m^2/kg^2, and v=9.0x10^-3 m/s:

`\begin{gathered} h=(1)/((1)/(6.37*10^6m)-((9.0*10^3\cdot(m)/(s))^2)/(2(6.67*10^(-11)N\cdot(m^2)/(kg^2))(5.98*10^(24)kg)))-6.37*10^6m \\ =18.03*10^6m-6.37*10^6m \\ =11.7*10^6m \end{gathered}`

Therefore, the maximum altitude reached by a rocket with an initial speed of 9.0x10^3m is:

`11.7*10^6m`