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?
(the polar radius of the earth is 6357 km)
You are wrong. The information given implies that this question is specifically testing the student's ability to take the height into account. The correct answer is 4.5 km/s. Could you please demonstrate how to obtain this answer?
My best guess is to determine the acceleration using a=GM/(r+h)^2 and then plug that acceleration into v^2=v0^2+a(y-y0). I had a similar problem in my homework and that got me a better answer than anything else I had tried.
To find the speed at which the object strikes the surface of the Earth, we can use the principle of conservation of mechanical energy. It states that the total mechanical energy of the object remains constant throughout its motion if we neglect any external forces acting on it. In this case, we are ignoring atmospheric friction, so we can use this principle.
The mechanical energy of the object consists of two components: the potential energy due to its altitude and the kinetic energy due to its speed. As the object falls, its potential energy decreases while its kinetic energy increases.
We can start by determining the potential energy of the object when it is at an altitude of 1200 km above the North Pole. The potential energy at any height is given by the formula:
Potential energy = mass * acceleration due to gravity * height
Here, the mass of the object is 910 kg, and the acceleration due to gravity at the surface of the Earth is approximately 9.8 m/s^2 (which remains relatively constant within the range considered).
However, we need to convert the height from kilometers to meters, so 1200 km becomes 1200000 m. Plugging the values into the formula, we get:
Potential energy = 910 kg * 9.8 m/s^2 * 1200000 m
Next, we need to determine the kinetic energy of the object just before it hits the surface of the Earth. The kinetic energy is given by the formula:
Kinetic energy = (1/2) * mass * velocity^2
Since the object starts from rest, its initial velocity is 0. We want to find the final velocity, which is the speed at which it strikes the Earth's surface. Plugging the mass (910 kg) and velocity (which is what we are solving for) into the formula, we get:
Kinetic energy = (1/2) * 910 kg * velocity^2
According to the principle of conservation of mechanical energy, the total mechanical energy remains constant. Therefore, the potential energy at the top should be equal to the kinetic energy at the bottom.
Now equate the potential energy and kinetic energy equations and solve for the velocity:
Potential energy = Kinetic energy
910 kg * 9.8 m/s^2 * 1200000 m = (1/2) * 910 kg * velocity^2
Simplifying the equation:
10584000000 J = (1/2) * 910 kg * velocity^2
Now solve for velocity:
velocity^2 = (2 * 10584000000 J) / 910 kg
velocity^2 = 23200000000 J/kg
velocity = sqrt(23200000000 J/kg)
Finally, calculate the square root of 23200000000 J/kg to find the velocity.
v = sqrt(2 g y) = sqrt(2*9.8*1200)
Unless they want to get picky about the change in g at that height.