Saved Bookmarks
| 1. |
A playful astronaut raleases a bowling ball, of mass m= 7.20 kg, into circular orbit about Earth at an altitude h of 350 km. (a) What is the mechanical energy E of the ball in its orbit ? (b) What is the mechanical energy E_0 of the ball on the launchpad at the Kennedy Space Center (bofore launch) ? From there to the orbit, what is the change DeltaE in the ball's mechanical energy? |
|
Answer» Solution :(a) We can get E from the orbital energy, given by Eq. 13-42 `(E = -GMm//2r)`, if we first find the orbital radius r. (It is not simply the given altitude.) Calculations : The orbital radius must be `r = R + H= 6370 km + 350 km= 6.72 xx 10^6m`, in which R is the radius of Earth. Then,from Eq.13-42 with Earth MASS `M = 5.98 xx 10^24kg `, the mechanical energy is `E = -(GMm)/(2r)` `=-((6.67 xx 10^(-11) N.m^2 //Kg^2 )(5.98 xx 10^24kg)(7.20kg))/((2)(6.72 xx 10^6m))` `=-2.14 xx 10^8 J = -214 MJ`. (b) On the launchpad, the ball is not in orbit and thus Eq. 13-42 does not apply. INSTEAD, we must find `E_0 = K_0 + U_0`, where `K_0` is the ball.s kenetic energy and `U_0` is the gravitational potential energy of the ball-Earth system. Calculations :To find `U_0`, we use Eq. 13-23 to write `U_0 = -(GMm)/(R)` `=-((6.67 xx 10^(-11)N.m^2//kg^2)(5.98xx10^24 kg)(7.20kg))/(6.37 xx 10^6m)` `=-4.51 xx10^8 J = -451 MJ`. The kinetic energy `K_0` of the ball is due to the ball.s MOTION with Earth.s rotation. You can show that `K_0` is less than 1 MJ,which is negligible relative to `U_0`. Thus, the mechanical energy of the ball on the launchpad is `E_0 = K_0 + U_0~~ 0-451 MJ = -451 MJ`. The increase in hte mechanical energy of the ball from launchpad to orbit is `DELTAE = E-E_0 = (-214 MJ) -(-451 MJ)` `=237 MJ`. This is worth a few dollars at your utility company. Obviously the high cost of placing objects into orbit is not due to their required mechanical energy. |
|