Final answer:
By applying the conservation of energy to an object falling towards a planet with no air resistance, the speed of the object before striking the surface is found by equating its initial potential energy to its final kinetic energy, resulting in an approximate velocity of 6.9 km/s. Option C is correct.
Step-by-step explanation:
The problem can be solved using the principle of conservation of energy which states that the total mechanical energy of an object (potential energy plus kinetic energy) remains constant if only conservative forces act upon the object. Since we are ignoring air resistance, we can use this principle to find the speed of the object just before it strikes the planet's surface.
The potential energy (PE) of the object when it's at height h above the surface is given by PE = GMm/(R+h), where G is the gravitational constant, M is the mass of the planet, m is the mass of the object, and R is the radius of the planet. The kinetic energy (KE) of the object just before striking the surface is KE = (1/2)mv², where v is the velocity we want to find. Because the object starts from rest, its initial kinetic energy is 0.
At the height h, the object has only potential energy, and just before it strikes the surface, it has converted all of this to kinetic energy. We equate the initial potential energy to the final kinetic energy to solve for v:
GMm/(R+h) = (1/2)mv²
By canceling m and solving for v, we get:
v = √(2GM/(R+h))
Substituting the given values:
v = √[2(6.674 × 10⁻¹¹ N·m²/kg²)(4.0 × 10²´ kg)/(5.0 × 10¶ m + 4.0 × 10¶ m)]
We calculate v and find it to be approximately 6.9 km/s, which corresponds to option C.