A meteor headed straight towards the earth has an approach speed of 2.00e3 m/s at a distance of 7 earth radii from the center of the earth. What will be its speed when it reaches earth a 6.38e6 m(ignore the effects of atmosphere)?
CAlculate the potential energy at each point:
PE= GMe*m/r
Then, set that change of energy into
original KE+ change of PE=final ke
Vf^2=Vi^2+ 2*change in PE/m
vf^2=vi^2+2GMe/r1-2GMe/r2
r1 original distance, r2 final
To find the speed of the meteor when it reaches the Earth, we can use the law of conservation of mechanical energy, which states that the sum of kinetic and potential energy is constant in the absence of external forces.
We know that the distance of the meteor from the center of the Earth is 7 Earth radii, which is equivalent to 7 times the radius of the Earth (6.38e6 m x 7 = 4.466e7 m). At this distance, the only energy that the meteor possesses is its kinetic energy.
The kinetic energy of an object is given by the formula:
Kinetic energy (K.E.) = (1/2)mv^2
Where m is the mass of the object and v is its speed.
To find the speed of the meteor when it reaches the Earth (v_final), we need to equate the initial kinetic energy (K.E. initial) to the final kinetic energy (K.E. final). Since no external forces act on the meteor, the total mechanical energy remains constant.
K.E. initial = K.E. final
Initially, when the meteor is 7 Earth radii away from the center of the Earth, its kinetic energy is given by:
K.E. initial = (1/2)m(initial) * (2.00e3 m/s)^2
Next, when the meteor reaches the Earth (6.38e6 m away from the center), its kinetic energy is given by:
K.E. final = (1/2)m(final) * v_final^2
Since the mass of the meteor will remain constant throughout its journey, we can simplify the equation:
K.E. initial = K.E. final
(1/2)m(initial) * (2.00e3 m/s)^2 = (1/2)m(final) * v_final^2
We can now solve for v_final:
v_final^2 = [(1/2)m(initial) * (2.00e3 m/s)^2] / (1/2)m(final)
v_final^2 = [(2.00e3 m/s)^2 * m(initial)] / m(final)
v_final^2 = (2.00e3 m/s)^2 * (m(initial) / m(final))
Finally, taking the square root of both sides will give us the speed of the meteor when it reaches the Earth:
v_final = √[(2.00e3 m/s)^2 * (m(initial) / m(final))]