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sergey [27]
2 years ago
5

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

at speed will an object hit the asteroid if it is dropped from 981.8 km above the surface?
Physics
1 answer:
makvit [3.9K]2 years ago
7 0

Answers:

(a) 2509.98 m/s

(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 700 km  and whose gravitational acceleration at the surface is a_{g}=4.5 m/s^{2}

Knowing this, let's begin:

a) In this part we need to find the escape speed V_{e} on the asteroid:

V_{e}=\sqrt{\frac{2GM}{R}} (1)

Where:

G is the universal gravitational constant

M is the mass of the asteroid

R=700 km=700(10)^{3} m is the radius of the asteroid

On the other hand we know the gravitational acceleration is a_{g}=4.5 m/s^{2}, which is given by:

a_{g}=\frac{GM}{R^{2}} (2)

Isolating GM:

GM=a_{g}R^{2} (3)

Substituting (3) in (1):

V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R} (4)

V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)} (5)

V_{e}=2509.98 m/s (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

E_{o}=E_{f} (7)

Being:

E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R} (8)

E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h} (9)

Where:

E_{o} is the initial mechanical energy

E_{f} is the final mechanical energy

K_{o} is the initial kinetic energy

K_{f}=0 is the final kinetic energy

U_{o} is the initial gravitational potential energy

U_{f} is the final gravitational potential energy

m is the mass of the object

V=1510 m/s is the radial speed of the object

h is the distance above the surface of the object

Then:

\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h} (10)

Isolating h:

h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R (11)

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 (11)

h=397042.215 m (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 h=981.8 km=981.8(10)^{3} m. This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2} (13)

Isolating V:

V=\sqrt{2a_{g} R(1-\frac{R}{R+h})} (14)

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})} (15)

Finally:

V=1917.76 m/s

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Explanation:

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\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

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