Respuesta :
Answers:
(a) [tex]2509.98 m/s[/tex]
(b) 397042.215 m
(c) 1917.76 m/s
Explanation:
The question is incomplete, please remember to write the whole question :) However, part (a) is written below:
(a) What is the escape speed on a spherical asteroid whose radius is [tex]700 km[/tex] and whose gravitational acceleration at the surface is [tex]a_{g}=4.5 m/s^{2}[/tex]
Knowing this, let's begin:
a) In this part we need to find the escape speed [tex]V_{e}[/tex] on the asteroid:
[tex]V_{e}=\sqrt{\frac{2GM}{R}}[/tex] (1)
Where:
[tex]G[/tex] is the universal gravitational constant
[tex]M[/tex] is the mass of the asteroid
[tex]R=700 km=700(10)^{3} m[/tex] is the radius of the asteroid
On the other hand we know the gravitational acceleration is [tex]a_{g}=4.5 m/s^{2}[/tex], which is given by:
[tex]a_{g}=\frac{GM}{R^{2}}[/tex] (2)
Isolating [tex]GM[/tex]:
[tex]GM=a_{g}R^{2}[/tex] (3)
Substituting (3) in (1):
[tex]V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}[/tex] (4)
[tex]V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}[/tex] (5)
[tex]V_{e}=2509.98 m/s[/tex] (6) This is the escape velocity
b) In this part we will use the Conservation of mechanical energy principle:
[tex]E_{o}=E_{f}[/tex] (7)
Being:
[tex]E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}[/tex] (8)
[tex]E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}[/tex] (9)
Where:
[tex]E_{o}[/tex] is the initial mechanical energy
[tex]E_{f}[/tex] is the final mechanical energy
[tex]K_{o}[/tex] is the initial kinetic energy
[tex]K_{f}=0[/tex] is the final kinetic energy
[tex]U_{o}[/tex] is the initial gravitational potential energy
[tex]U_{f}[/tex] is the final gravitational potential energy
[tex]m[/tex] is the mass of the object
[tex]V=1510 m/s[/tex] is the radial speed of the object
[tex]h[/tex] is the distance above the surface of the object
Then:
[tex]\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}[/tex] (10)
Isolating [tex]h[/tex]:
[tex]h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R[/tex] (11)
[tex]h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m[/tex] (11)
[tex]h=397042.215 m[/tex] (12) This is the distance above the asteroid's surface
c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance [tex]h=981.8 km=981.8(10)^{3} m[/tex]. This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:
[tex]\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}[/tex] (13)
Isolating [tex]V[/tex]:
[tex]V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}[/tex] (14)
[tex]V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}[/tex] (15)
Finally:
[tex]V=1917.76 m/s[/tex]
