Question

(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface is 3.00 m/s2 ? (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1000 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface?

Answer 1

The escape velocity of a spherical asteroid with a given radius and given gravitational acceleration can be calculated using a special formula. The distance a particle travels from the surface when it leaves it at a given radial velocity can be determined using the projectile height equation. The speed at which an object hits an asteroid after falling from a given height can be calculated using the falling object terminal velocity equation.

Step-by-step explanation:

(a) Escape velocity can be defined as the minimum speed required for an object to escape the gravitational pull of a celestial body. To calculate the escape velocity on a spherical asteroid we can use the formula:

escape velocity = sqrt(2 * acceleration due to gravity * radius).

Using the given values, the escape velocity of a spherical asteroid is:

exhaust rate = sqrt(2 * 3.00 m/s2 * 500 000 m) = 6928 m/s.

(b) To calculate the distance from the surface that a particle will travel when it leaves the surface of the asteroid with a radial velocity of 1000 m/s, we can use the formula for the height of the projectile:

Height = (radial velocity)2 / (2 * acceleration due to gravity).

After entering the specified values, the particle travels the distance:

Altezza = (1.000 m/s)2 / (2 * 3,00 m/s2) = 166.666,67 m.

(c) To calculate the speed at which an object hits an asteroid as it falls 1000 km above the surface, we can use the equation for the final velocity of the falling object:

Final velocity = sqrt (initial velocity2 + 2 * acceleration due to gravity * height).

Substituting the given values, the object hits the asteroid with a speed of:

Final velocity = square(0 + 2 * 3.00 m/s2 * 1,000,000 m) = square(6,000,000 m2/s2) = 2,449 m/s.

Answer 2

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Step-by-step explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be
`U_i =-(GMm)/(R)`.

The initial kinetic energy would be
`(1)/(2)mv^2`. The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

`-(GMm)/(R)+(1)/(2)mv^2=0` (1)

Dividing both sides by m and replacing
`(GM)/(R)` by
`a_g R`

And the equation (1) becomes:

`-a_g R+(1)/(2) v^2=0` (2)

If we solve for v we got this:

`v=√(2 a_g R)=\sqrt{2x3(m)/(s^2)x500000m}=1732.05m/s`

Part b

When we consider a particule at this surface at the starting point we have that:

`U_i=-(GMm)/(R)`

`K_i=(1)/(2)mv^2`

Considering that the particle is at a distance h above the surface and then stops we have that:

`U_f=-(GMm)/(R+h)`

`K_f=0`

And the balance of energy would be:

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

Dividing again both sides by m and replacing
`(GM)/(R)` by
`a_g R^2` we got:

`-a_g R+(1)/(2)v^2 =-(a_g R^2)/(R+h)`

If we solve for h we can follow the following steps:

`R+h=-(a_g R^2)/(-a_g R+(1)/(2)v^2)`

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

`h=(2a_g R^2)/(2a_g R-v^2)-R`

Replacing:

`h=(2x3(m)/(s^2)(500000m)^2)/(2(3(m)/(s^2))(500000m)-(1000m/s)^2)- 500000m=250000m`

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

`U_i=-(GMm)/(R+h)`

`K_i=0`

And after the particle reach the asteroid we have this:

`U_f=-(GMm)/(R)`

`K_f=(1)/(2)mv^2`

So the balance of energy would be:

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

Replacing again
`a_g R^2` instead of GM and dividing both sides by m we have:

`-(a_g R^2)/(R+h)=-a_g R+(1)/(2)v^2`

And solving for v:

`a_g R-(a_g R^2)/(R+h)=(1)/(2)v^2`

Multiplying both sides by two and taking square root:

`v=\sqrt{2a_g R-(2a_g R^2)/(R+h)}`

Replacing

`v=\sqrt{2(3(m)/(s^2))(500000m)-(2(3(m)/(s^2)(500000m)^2)/(500000+1000000m)}=1414.214m/s`