What is the launch speed of a projectile that rises vertically above the Earth t

Question

What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to one Earth's radius before coming to rest momentarily? (Astronomical data needed for this problem can be found on the inside back cover of the text. Ignore air resistance.) km/s

Explanation / Answer

PLEASE RATE ME AND AWARD ME KARMA POINTS IF IT IS HELPFUL FOR YOU The problem with the other solution is that they assumed the accleration of gravity is constant over the distances involved. Although this is a valid assumption for small displacements close to the surface of the Earth it is not true when larger distances are involved. g = GM/Re^2 ..... acceleration of gravity at the surface of the Earth a = GM/R^2 ....... acceleration of gravity at some distance R from the center of the Earth a/g = (Re/R)^2 = 1/4 since R = 2*Re So clearly the accleration of gravity is anything but constant over the distances involved. If you assume that it is constant it is equally clear that you will arrive at a velocity that is too large. To find the launch speed we will use the conservation of energy. The kinetic energy at launch plus the potential energy at launch must equal the potential energy at altitude since the projectile comes to rest at this altitude. M = mass of the Earth = 5.9742x10^24 kg G = gravitational constant = 6.673x10^(-11) m^3 / (kg s^2) Re = radius of the Earth = 6.378x10^6 m R = max height of the projectile = 2*Re = 1.2756x10^6 m m = mass of the projectile PEa = PE at altitude = -GMm/R PEe = PE at Earths surfact = -GMm/Re KE = kinetic energy at launch KE + PEe = PEa .... conservation of energy KE = PEa - PEe (1/2)mv^2 = GMm[1/Re - 1/R] v^2 = 2*GM*[1/Re - 1/R] = GM/Re since R = 2*Re v = SQRT[G*M/Re] = 7906 m/s = 7.906 km/s You can see that this method has a limit and that would be when R -> Infinity in which case you have for initial velocity: v = SQRT[2*GM/Re] And this is called the escape velocity and is equal to 11,2 km/s Also this is exactly what that other answer calculated. g = GM/Re^2 so v = SQRT(2*g*Re) But the problem wants to know the initial speed to launch a projectile to an altitude equal to the radius of the Earth and not the escape velocity. I suppose I should also point out that the other answer is not the escape velocity since it is dependent on altitude of the projectile and in fact goes to infinity as that altitude goes to infinity. It is only a coincidence that it worked out as it did. Some other altitude and it would no

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.