A 910kg object is released from rest at an altitude of 1200 km above the north pole of the earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth?
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To find the speed at which the object strikes the surface of the Earth, we can use the concept of gravitational potential energy and the conservation of mechanical energy.
The gravitational potential energy (PE) of an object at a height above the Earth's surface is given by the equation:
PE = mgh
where m is the mass of the object, g is the acceleration due to gravity, and h is the height above the Earth's surface.
In this case, the object is released from rest, so initially, it has no kinetic energy (KE). As the object falls towards the Earth, its potential energy is converted into kinetic energy. According to the conservation of mechanical energy, the total mechanical energy (ME) of the object remains constant throughout its fall.
ME = PE + KE
At the surface of the Earth, when the object strikes the ground, its potential energy is zero and all the mechanical energy is in the form of kinetic energy.
KE = ME
Let's break down the solution step by step:
Step 1: Calculate the gravitational potential energy (PE) at an altitude of 1200 km:
PE = mgh
Given: m = 910 kg, h = 1200 km = 1200000 m (convert km to m), g = 9.8 m/s² (acceleration due to gravity)
PE = 910 kg × 9.8 m/s² × 1200000 m
Step 2: Calculate the initial mechanical energy (ME) of the object:
ME = PE
Step 3: Calculate the final kinetic energy (KE) of the object when it strikes the surface of the Earth:
KE = ME
Step 4: Calculate the final speed (v) of the object:
KE = 0.5mv² (kinetic energy formula)
Solve the equation for v:
0.5mv² = KE
0.5 × 910 kg × v² = ME
v² = (2 × ME) / m
v = √[(2 × ME) / m]
Substitute the values of ME and m to find the speed v.
Remember to convert units if needed.
By following these steps and plugging in the given values, you can find the speed at which the object strikes the surface of the Earth.