Question

object of mass \(m\) is launched from the surface of the Earth. Show that the minimum speed required to send the projectile to a height of \(4 R_{\mathrm{E}}\) above the surface of the Earth is \(v_{\min }=\sqrt{8 G M_{\mathrm{E}} / 5 R_{\mathrm{E}}} \cdot M_{\mathrm{E}}\) is the mass of the Earth and \(R_{\mathrm{E}}\) is the radius of the Earth. Neglect air resistance.

Short answer

Answer: The minimum speed required to send the object to a height of \(4R_E\) above the surface of the Earth is \(v_{min} = \sqrt{\frac{8GM_E}{5R_E}}\).

Step-by-step solution

Write down the conservation of energy equation

Since we are neglecting air resistance, we can use conservation of mechanical energy, which states that the total initial energy equals the total final energy. Total initial energy (at the surface) = Total final energy (at a height of \(4R_E\)) \(E_{initial} = E_{final}\)

Identify the forms of energy in the initial and final states

Initially, the object has both kinetic energy \(K_{initial}\) and gravitational potential energy \(U_{initial}\). At the final height of \(4R_E\), the object has gravitational potential energy \(U_{final}\). Since we are considering minimum velocity, it will have no kinetic energy at the final height. \(K_{initial} + U_{initial} = U_{final}\)

Write down the equations for kinetic energy and gravitational potential energy

We can write down the expressions for the initial kinetic energy and gravitational potential energies at both the initial and final positions: \(K_{initial} = \frac{1}{2}mv_{min}^2\) \(U_{initial} = -\frac{GM_Em}{R_E}\) \(U_{final} = -\frac{GM_Em}{5R_E}\) Now substitute these expressions back into the energy conservation equation from Step 2: \(\frac{1}{2}mv_{min}^2 - \frac{GM_Em}{R_E} = -\frac{GM_Em}{5R_E}\)

Solve for the minimum velocity \(v_{min}\)

Now we can solve for \(v_{min}\): \(\frac{1}{2}mv_{min}^2 = \frac{GM_Em}{R_E} - \frac{GM_Em}{5R_E}\) \(v_{min}^2 = \frac{2GM_Em}{R_E} - \frac{2GM_Em}{5R_E}\) Factor out \(2GM_Em\): \(v_{min}^2 = 2GM_Em\left(\frac{1}{R_E} - \frac{1}{5R_E}\right)\) \(v_{min}^2 = \frac{8GM_Em}{5R_E}\) \(v_{min} = \sqrt{\frac{8GM_E}{5R_E}}\) The minimum speed required to send the object to a height of \(4R_E\) above the surface of the Earth is: \(v_{min} = \sqrt{\frac{8GM_E}{5R_E}}\)