Question

In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid the size of a small city. The previously unknown asteroid, now designated 2002 MN, remained undetected until three days after it had passed the Earth. Suppose that at its closest approach, the asteroid was 75900 miles from the center of the Earth -- about a third of the distance to the Moon. (a) Find the speed of the asteroid at closest approach, assuming its speed at infinite distance to be zero and considering only its interaction with the Earth. km/s (b) Observations indicate the asteroid to have a diameter of about 2.0 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it has an average density of 3.33 g/cm3. (For comparison, a 1-megaton nuclear weapon releases about 5.6 multiplied by 1015 J of energy.)

Answer 1

2.55383611 km/s

`4.54872* 10^(19)\ J`

Step-by-step explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth = 5.972 × 10²⁴ kg

`\rho` = Density of asteroid = 3.33 g/cm³ = 3330 kg/m³

r = Distance between asteroid and Earth = 75900 miles

D = Diameter of asteroid = 2 km

R = Radius =
`(D)/(2)=1\ km`

V = Volume =
`(4)/(3)\pi R^3` (assumed shape is a sphere)

Converting to m

`1\ mile=1609.34\ m`

`75900\ mile=75900* 1609.34\ m=122148906\ m`

The potential and kinetic energies are conserved

`P_i=K_f+P_f\\\Rightarrow 0=(1)/(2)mv^2-(GMm)/(r)\\\Rightarrow v=\sqrt{(2GM)/(r)}\\\Rightarrow v=\sqrt{(2* 6.67* 10^(-11)* 5.972* 10^(24))/(122148906)}\\\Rightarrow v=2553.83611\ m/s=2.55383611\ km/s`

The speed of the asteroid at closest approach is 2.55383611 km/s

Kinetic energy is given by

`K=(1)/(2)mv^2\\\Rightarrow K=(1)/(2)* \rho* V* v^2\\\Rightarrow K=(1)/(2)* 3330* (4)/(3)\pi 1000^3* 2553.83611^2\\\Rightarrow K=4.54872* 10^(19)\ J`

The kinetic energy of the asteroid is
`4.54872* 10^(19)\ J`