Saved Bookmarks
| 1. |
A spaceship of mass m = 4.50 xx 10^3 kg is in a circular Earth orbit of radiusr = 8.00 xx 10^6m and period T_(0) = 118.6 min=7.119 xx 10^3 s when a thruster is fired in the forward direction to decrease the speed to 96.0% of the original speed.What is the period T of the resulting elliptical orbit ? |
Answer» Solution :(1) An ellipical orbit period is related to the semimajor exis a by Kepler.s THIRD law,written as Eq. 13.36 `(T^2 = 4pi^2r^3//GM)` but withareplacing r.(2) The semima-jor axis is related to the total mechanical energy E of the ship by Eq. 13-44 (E = -GMm/2a), in which Earth.s  mass is `M = 5.98 xx 10^24 kg`. (3) The potential energy of the ship at radius r from Earth.s center is given by Eq. 13-23 (U = -GMm/r). Calculations : Looking over the Key Ideas, we see that we need to CALCULATED the total energy E to find the semimajor axis a, so that we can then determine the period of the elliptical orbit. Let.s start with the KINETIC energy, calculating it just after the thruster is fired. The speed v just then is 96% of the circumference of the initial circular orbit tot he initial period of the orbit. Thus, just after the thrusteris fired, the kineticenergy is `K = 1/2mv^2 = 1/2 m(0.96v_0)^2 = 1/2 m (0.96)^2 ((2pir)/(T_0))^2` `=1/2 (4.50 xx 10^3 kg) (0.96)^1 ((2pi(8.00xx10^6m))/(7.119xx10^3s))^2` `=1.0338 xx 10^11J`. Just after the thruster is fired,the ship is still at orbital radius r, and thus its gravitational protential energy is `U = -(GMm)/(r)` `=-((6.67xx10^(-11)N.m^2//kg^2)(5.98xx10^24kg)(4.50xx10^3kg))/(8.00xx10^6m)` `=-2.2436 xx 10^11 J`. We can now find the semimojar axis by REARRANGING Eq. 13.44, substituting a for , and then substitutingin our energy results: `a = -(GMm)/(2E) = -(GMm)/(2(K+U))` `=-((6.67xx10^(-11)N.m^2//kg^2)(5.98xx10^24kg)(4.50xx10^3kg))/(2(1.0338xx10^11J-2.2436xx10^11J))` `= 7.418 xx 10^6m`. Now, one more step to go. We substitute a for r in Eq. 13.36 and then solve for the period T,substituting our result for a: `T = ((4pi^2a^3)/(GM))^(1//2)` `=((4pi^2(7.418xx10^6m)^3)/((6.67xx10^(-11)N.m^2//kg^2)(5.98xx10^24kg)))^(1//2)` `=6.356xx10^3 s = 106 min`. This is the period of the elliptical orbit that the ship takes after the thrusteris fired.It is less than the period `T_0` for the circular orbti for two reasons. (1) The orbital path length is now less. (2) The elliptical path takes the ship closer to Earth everywhere except at the POINT of firing. The resulting decrease in gravitational potential energy increases the kinetic energy and thus also the speed of the ship. |
|