Question

A projectile is launched straight upward from Earth's surface at the North Pole. For Earth's radius at the North Pole, use

Answer 1

Answer:

a) Maximum altitude of the projectile, $h = 2.83 * 10^5 m$

b) v = 10885.7 m/s

Explanation:

a) Mass of the earth, $m_{1} = 5.98 * 10^{24} kg$

Mass of the projectile = $m_{2}$

Launch speed, v = 2.31 * 10³ m/s

Earth radius, r =  6.36 ✕ 10⁶ m

Workdone by the projectile against gravity

$W = Gm_{1} m_{2} (\frac{1}{r} - \frac{1}{r+h} )\\W = 5.98 * 10^{24} *6.67 * 10^{-11} m_{2} (\frac{1}{6.36*10^{6} } - \frac{1}{r+h} )$

$W = 398866 * 10^{9} m_{2}(\frac{1}{6.36*10^{6} } - \frac{1}{r+h} )$...............(1)

Kinetic energy of the projectile:

$KE = \frac{1}{2} m_{2} v^{2}$

$KE =2668050 m_{2}$...................(2)

Equating (1) and (2) based on the law of energy conservation

$398866 * 10^{9} m_{2}(\frac{1}{6.36*10^{6} } - \frac{1}{r+h} ) = 2668050 m_{2}\\\frac{1}{6.36*10^{6} } - \frac{1}{r+h} = 0.00000000669\\0.00000015723 - 0.00000000669 = \frac{1}{r+h}\\0.00000015054 = \frac{1}{(6.36*10^{6}) +h}\\(6.36*10^{6}) +h = 6642633.424\\h = 6642633.424 - (6.36*10^{6})\\h = 282633.42 m\\h = 2.83 * 10^5 m$

b) Smallest required change in the satellite speed

Altitude, h = 352 km = 352000 m

Earth radius, r = 6.38 * 10⁶ m

$v = \sqrt{\frac{2Gm_{1} }{(r+h)} } \\v = \sqrt{\frac{2* 6.67 * 10^{-11} *5.98*10^{24} }{(6.38*10^6+352*10^3)} }$

v = 10885.7 m/s