Question

(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 what speed will an object hit the asteroid if it is dropped from 981.8 km above the surface?

Answer 1

A particle leaving the asteroid's surface with a radial speed of 1510 m/s will go approximately 8.39 km from the surface. An object dropped from 981.8 km above the surface will hit the asteroid with a velocity of approximately 4.72 km/s.

Step-by-step explanation:

To find how far from the surface a particle will go when it leaves the asteroid's surface with a radial speed of 1510 m/s, we can use the conservation of energy. The potential energy at the surface will be converted to kinetic energy. We can calculate the distance by considering the change in potential energy and equating it to the change in kinetic energy. Using this method, we find that the particle will go a distance of approximately 8.39 km from the surface.

For the second question, when an object is dropped from 981.8 km above the surface, it will gain potential energy as it falls towards the asteroid. This potential energy will be converted to kinetic energy just before it hits the surface. Using the conservation of energy, we can calculate the velocity of the object just before it hits the surface. Using this method, we find that the object will hit the asteroid's surface with a velocity of approximately 4.72 km/s.

Answer 2

Answers:

(a)
`2509.98 m/s`

(b) 397042.215 m

(c) 1917.76 m/s

Step-by-step 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{(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)=(GM)/(R^(2))` (2)

Isolating
`GM`:

`GM=a_(g)R^(2)` (3)

Substituting (3) in (1):

`V_(e)=\sqrt{(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)=(1)/(2)m V^(2) - (GMm)/(R)` (8)

`E_(f)=K_(f)+U_(f)=0 - (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:

`(1)/(2)m V^(2) - (GMm)/(R)=- (GMm)/(R+h)` (10)

Isolating
`h`:

`h=(2 a_(g) R^(2))/(2a_(g)R-V^(2))-R` (11)

`h=(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:

`(-GMm)/(R+h)=(-GMm)/(R)+(1)/(2)mV^(2)` (13)

Isolating
`V`:

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

`V=\sqrt{2(4.5 m/s^(2)) (700(10)^(3) m)(1-(700(10)^(3) m)/(700(10)^(3) m+981.8(10)^(3) m))}` (15)

Finally:

`V=1917.76 m/s`