Question

15. An asteroid with mass m = 1.00 x 10 kg comes from deep space, effectively from infinity, and falls toward Earth. (a) Find the change in potential energy when it reaches a point 4.00 x 108 m from Earth (just beyond the Moon), assuming it falls from rest at infinity. In addition, find the work done by the force of gravity. (b) Calculate the speed of the asteroid at that point. (c) How much work would have to be done on the asteroid by some other agent so the asteroid would be traveling at only half the speed found in (b) at the same point?​

Answer 1

To solve this problem, we can use the concepts of gravitational potential energy, work, and conservation of mechanical energy.

Let's break down the problem step by step:

Given data:

Mass of the asteroid, m = 1.00 x 10^5 kg

Initial distance from Earth (infinity), r_initial = ∞

Final distance from Earth, r_final = 4.00 x 10^8 m (just beyond the Moon)

Gravitational constant, G = 6.67430 x 10^-11 m^3 kg^-1 s^-2

Mass of Earth, M = 5.972 x 10^24 kg

(a) Change in Potential Energy and Work Done by Gravity:

The change in potential energy ΔU of the asteroid as it falls from infinity to a distance of r_final is given by the formula:

ΔU = -G * M * m * (1/r_final - 1/r_initial)

Since the asteroid starts from infinity, its initial potential energy is considered zero (U_initial = 0). Therefore, the work done by gravity is equal to the change in potential energy:

Work = ΔU

Substitute the values:

ΔU = -G * M * m * (1/r_final)

= -6.67430 x 10^-11 m^3 kg^-1 s^-2 * 5.972 x 10^24 kg * 1.00 x 10^5 kg * (1 / 4.00 x 10^8 m)

≈ -1.252 x 10^9 joules

So, the change in potential energy (and work done by gravity) is approximately -1.252 x 10^9 joules.

(b) Speed of the Asteroid:

The conservation of mechanical energy states that the sum of an object's kinetic energy (KE) and potential energy (PE) remains constant if only conservative forces (like gravity) are acting on it. Therefore:

Initial mechanical energy (E_initial) = Final mechanical energy (E_final)

At infinity, the asteroid's kinetic energy is zero, so its mechanical energy is equal to its potential energy:

E_initial = U_initial = 0

At a distance of r_final, the mechanical energy is the sum of kinetic and potential energy:

E_final = KE_final + PE_final

Since the asteroid's potential energy at infinity is zero, the mechanical energy conservation equation becomes:

0 = 0.5 * m * v_final^2 - G * M * m / r_final

Solving for v_final (speed of the asteroid):

v_final^2 = 2 * G * M / r_final

v_final = √(2 * G * M / r_final)

Substitute the values:

v_final = √(2 * 6.67430 x 10^-11 m^3 kg^-1 s^-2 * 5.972 x 10^24 kg / 4.00 x 10^8 m)

≈ 1.015 x 10^3 m/s

The speed of the asteroid at that point is approximately 1.015 x 10^3 m/s.

(c) Work to Change Speed:

The work done by an external agent to change the speed of the asteroid from v_final to half its value is equal to the change in kinetic energy:

Work_external = ΔKE = 0.5 * m * (v_final/2)^2 - 0.5 * m * v_final^2

Substitute the values:

Work_external = 0.5 * 1.00 x 10^5 kg * (1.015 x 10^3 m/s / 2)^2 - 0.5 * 1.00 x 10^5 kg * (1.015 x 10^3 m/s)^2

≈ -1.281 x 10^8 joules

The work done by the external agent to reduce the asteroid's speed to half its value is approximately -1.281 x 10^8 joules.

[The negative sign indicates that work is done on the asteroid by the external agent to reduce its speed.]