Respuesta :
Answer: [tex]V_{1}<V_{3}<V_{2}[/tex]
a) [tex]V_{1}=1.317(10)^{-9} m/s[/tex]
b) [tex]V_{2}=1.02(10)^{-8} m/s[/tex]
c) [tex]V_{3}=1.61(10)^{-9} m/s[/tex]
Explanation:
Assuming these objects are moving with uniform circular motion (and the orbit is perfectly circular), the velocity will be given by the following equation:
[tex]V=\sqrt{G\frac{m}{r}}[/tex]
Where:
[tex]V[/tex] is the velocity of the object, which is assumed as constant.
[tex]G=6.67408(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant
[tex]m[/tex] is the mass of the object
[tex]r[/tex] is the radius of the orbit
Knowing this, let's begin with the answers:
a. a 5-kg object orbiting Earth halfway to the moon
In this case the radius of the orbit is the half of the distance between the Earth and the moon:
[tex]r_{1}=\frac{d_{Earth-Moon}}{2}=\frac{384400000 m}{2}[/tex]
[tex]r_{1}=192200000 m[/tex]
And the mass will be [tex]m=5 kg[/tex]. So, the equation of the velocity is:
[tex]V_{1}=\sqrt{G\frac{m_{1}}{r_{1}}}[/tex]
[tex]V_{1}=\sqrt{6.67408(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5 kg}{192200000 m}}[/tex]
[tex]V_{1}=0.000000001317 m/s=1.317(10)^{-9} m/s[/tex] This is the velocity of the first object
b. a 10-kg object orbiting Earth just above Earth's surface
Now the radius of the orbit is approximately the radius of the Earth:
[tex]r_{2}=r_{Earth}=6371000 m[/tex]
And the mass is [tex]m_{2}=10 kg[/tex]. So, the equation of the velocity is:
[tex]V_{2}=\sqrt{G\frac{m_{2}}{r_{2}}}[/tex]
[tex]V_{2}=\sqrt{6.67408(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{10 kg}{6371000 m}}[/tex]
[tex]V_{2}=0.0000000102 m/s=1.02(10)^{-8} m/s[/tex] This is the velocity of the second object
c. a 15-kg object orbiting Earth at the same distance as the Moon
With this last case, the radius of the orbit is equal to the distance between the Earth and the Moon:
[tex]r_{3}=r_{Earth-Moon}=384400000 m[/tex]
And the mass is [tex]m_{3}=15 kg[/tex]. So, the equation of the velocity is:
[tex]V_{3}=\sqrt{G\frac{m_{3}}{r_{3}}}[/tex]
[tex]V_{3}=\sqrt{6.67408(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{15 kg}{384400000 m}}[/tex]
[tex]V_{3}=0.00000000161 m/s=1.61(10)^{-9} m/s[/tex] This is the velocity of the third object
Finally, comparing the velocity of the three objects we have:
[tex]1.317(10)^{-9} m/s < 1.61(10)^{-9} m/s < 1.02(10)^{-8} m/s[/tex]
[tex]V_{1} <V_{3} < V_{2}[/tex]