A meteoroid is moving towards a planet. It has mass m= 0.18x10⁹ kg and speed v1 = 5.9x10⁷ m/s at distance R1 = 2.9x10⁷ m from the center of the planet. The radius of the planet is R = 0.22x10⁷ m. The mass of the planet is M = 2.4x10²⁵ kg. There is no air around the planet.
Calculate the speed of the meteoroid just prior to the instant when it hits the surface of the planet, in meters per second.

5.90000114e+7 m/s

Explanation:

Given values

M₁ = 1.8e+8 kg

M₂ = 2.4e+25 kg

M = M₁ + M₂ = 2.400000000000000018e+25 kg

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

GM = 1.60183200000000001201374e+15 m³ sec⁻²

r₁ = 2.9e+7 m

v₁ = 5.9e+7 m/s

R = 2.2e+6 m

The total energy of the meteroid per unit mass at time t₁ is

E/M₁ = 0.5v₁² − GM/r₁ = 1.7404999447644137931034478615952e+15 m²/s²

At impact,

E/M₁ = 0.5v₂² − GM/R = 1.7404999447644137931034478615952e+15 m²/s²

v₂ = √[2(E/M₁ + GM/R)] = 5.9000011404572937396882314817886e+7 m/s

v₂ − v₁ ≈ 11.4 m/s

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~