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A projectile is launched straight upward from Earth's surface at the North Pole. For Earth's radius at the North Pole, use
RE, NP = 6.36 ✕ 106 m.
(a) Ignoring air resistance, find the maximum altitude of the projectile above Earth's surface if the launch speed is
2.31 ✕ 103 m/s. _____m

A satellite is in a circular low Earth orbit at an altitude of
352 km
above the equator's surface. For Earth's radius at the equator, use
RE, eq = 6.38 ✕ 106 m.
(b) Assuming only the force of Earth's gravity acts on the satellite, determine the smallest required change in the satellite's speed if it is to escape Earth's gravity and never return. _____m/s

Respuesta :

Answer:

a) Maximum altitude of the projectile, [tex]h = 2.83 * 10^5 m[/tex]

b) v = 10885.7 m/s

Explanation:

a) Mass of the earth, [tex]m_{1} = 5.98 * 10^{24} kg[/tex]

Mass of the projectile = [tex]m_{2}[/tex]

Launch speed, v = 2.31 * 10³ m/s

Earth radius, r =  6.36 ✕ 10⁶ m

Workdone by the projectile against gravity

[tex]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} )\\[/tex]

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

Kinetic energy of the projectile:

[tex]KE = \frac{1}{2} m_{2} v^{2}[/tex]

[tex]KE =2668050 m_{2}[/tex]...................(2)

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

[tex]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[/tex]

b) Smallest required change in the satellite speed

Altitude, h = 352 km = 352000 m

Earth radius, r = 6.38 * 10⁶ m

[tex]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)} }[/tex]

v = 10885.7 m/s

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