Answer:
a).[tex]v_{1}=13.49 \frac{m}{s}[/tex]
b).[tex]v_{2}=17.54\frac{m}{s}[/tex]
Explanation:
Using the conservation of energy and the kinetic energy of the satellite around the asteroid can model the motion in
[tex]\frac{1}{2}*m*v^2=\frac{G*M*m}{a_{o}}[/tex]
[tex]v^2=\frac{2*G*M*m}{a_{o}}[/tex]
[tex]G=6.673x10^{-11}\frac{m^3}{kg*s^2}[/tex]
[tex]M=1.20x10^{16}kg[/tex]
a).
[tex]a_{o}=8.8km*\frac{1000m}{1km}=8800m[/tex]
[tex]v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{8800m}}[/tex]
[tex]v_{1}=13.49 \frac{m}{s}[/tex]
b).
[tex]a_{f}=5.2km*\frac{1000m}{1km}=5200m[/tex]
[tex]v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{5200m}}[/tex]
[tex]v_{2}=17.54\frac{m}{s}[/tex]
Answer:
a). What is the speed of a satellite orbiting 5.20km above the surface?
The speed of the satellite is [tex]4.88m/s[/tex].
b). What is the escape speed from the asteroid?
The escape speed from the asteroid is [tex]6.90m/s[/tex].
Explanation:
a). What is the speed of a satellite orbiting 5.20km above the surface?
The speed of the satellite can be found through the Universal law of gravity:
[tex]F = G\frac{M \cdot m}{r^{2}}[/tex] (1)
Then, replacing Newton's second law in equation 1 it is gotten:
[tex]m.a = G\frac{M \cdot m}{r^{2}}[/tex] (2)
However, a is the centripetal acceleration since the satellite describes a circular motion around the asteroid:
[tex]a = \frac{v^{2}}{r}[/tex] (3)
Replacing equation 3 in equation 2 it is gotten:
[tex]m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}[/tex]
[tex]m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r[/tex]
[tex]m \cdot v^{2} = G \frac{M \cdot m}{r}[/tex]
[tex]v^{2} = G \frac{M \cdot m}{rm}[/tex]
[tex]v^{2} = G \frac{M}{r}[/tex]
[tex]v = \sqrt{\frac{G M}{r}}[/tex] (4)
Where v is the orbital speed, G is the gravitational constant, M is the mass of the asteroid, and r is the orbital radius.
Notice that the orbital radius will be given by the sum of the radius of the asteroid and the height of the satellite above the surface:
[tex]r = 8.80km+5.20km[/tex]
[tex]r = 14km[/tex]
But it is necessary to express r in units of meters before it can be used in equation 4.
[tex]r = 14km \cdot \frac{1000m}{1km}[/tex] ⇒ [tex]14000m[/tex]
[tex]v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(5.00x10^{15}kg)}{14000m}}[/tex]
[tex]v = 4.88m/s[/tex]
Hence, the speed of the satellite is [tex]4.88m/s[/tex]
b). What is the escape speed from the asteroid?
The escape speed is defined as:
[tex]v_{e} = \sqrt{\frac{2GM}{r}}[/tex] (5)
Where G is the gravitational constant, M is the asteroid radius and r is the orbital radius.
[tex]v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.00x10^{15}kg)}{14000m}}[/tex]
[tex]v_{e} = 6.90m/s[/tex]
Hence, the escape speed from the asteroid is [tex]6.90m/s[/tex].
