A 1600 kg object is initially at rest 400 km above the earth’s surface.

In summary, the conversation discusses a problem involving an object falling from a height of 400 km above the Earth's surface and generating heat energy. The solution involves using equations for kinetic and potential energy, as well as the force of gravity, to calculate the velocity of the object at an altitude of 100 km. A mistake is found in the calculation, which is corrected by converting kilometers to meters.
  • #1
Scarlett
3
0

Homework Statement



A 1600 kg object is initially at rest 400 km above the earth’s surface. The object falls straight
down and generates 8.3x10^8J of heat energy while descending to an altitude of 100 km.
What is the velocity of the object at this altitude?
(http://i1070.photobucket.com/albums/u488/ScatlettArmadillo/Question20.jpg)

Homework Equations



Ek = (mv^2)/2
Ep = -GMm/r
G=6.67^-11
Mass of Earth = 5.98x10^24 kg
Radius of Earth = 6380000 m
Force of Gravity = GMm/r^2

The Attempt at a Solution



http://i1070.photobucket.com/albums/u488/ScatlettArmadillo/Question20-1.jpg
Once I got to the final step I had to find the root of a negative, which is not possible, so I must have made an error somewhere else, but I can't seem to work out where.
 
Last edited:
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  • #2
Your solution is correct, but the mistake lies in 1/r2 - 1/r1, this equals 6x10^-9, not 7.37 x 10^-12.
 
  • #3
Xisune said:
Your solution is correct, but the mistake lies in 1/r2 - 1/r1, this equals 6x10^-9, not 7.37 x 10^-12.

How?
(1/(6380000+100))-(1/(6380000+400))
=(1/6380100)-(1/6380400)
=1.57x10^-7-1.58x10-7
=7.37x10^-12
 
  • #4
Oh, god, I knew I'd've made a stupid mistake, I forgot to convert km to m. Ugh, stupid, but at least I got it now, thanks for the help :)
 
  • #5


I would approach this problem by first checking my calculations and equations to make sure they are correct. It seems that you have made a mistake in your calculation of the force of gravity. The correct equation for force of gravity is F = GMm/r^2. You have used r instead of r^2 in your calculation.

Once you have corrected this mistake and recalculated the force of gravity, you can use the equation F = ma to find the acceleration of the object. From there, you can use the equation v^2 = u^2 + 2as to find the final velocity of the object at an altitude of 100 km. This should give you a positive value for the velocity, as it is impossible to have a negative velocity.

Additionally, as a scientist, I would also consider the assumptions and limitations of this problem. For example, the object is assumed to be falling in a vacuum, neglecting air resistance. This may affect the accuracy of the calculations. I would also consider the potential for energy loss due to other factors, such as friction or heat transfer to the surroundings.
 

1. What is the initial potential energy of the object at rest 400 km above the earth's surface?

The initial potential energy of the object can be calculated using the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s²), and h is the height above the surface. Plugging in the values, we get PE = (1600 kg)(9.8 m/s²)(400,000 m) = 6.24 x 10^9 J.

2. How much work is required to move the object from 400 km above the earth's surface to the surface itself?

The work done to move the object from 400 km above the surface to the surface can be calculated using the formula W = Fd, where F is the force applied and d is the distance moved. Since the object is being moved against the force of gravity, the work done would be equal to the change in potential energy, which we calculated in the previous question. Therefore, the work done would also be 6.24 x 10^9 J.

3. What is the final velocity of the object when it reaches the surface of the earth?

Using the law of conservation of energy, we can equate the initial potential energy to the final kinetic energy of the object. Thus, PE = KE, or mgh = 1/2mv^2. Solving for v, we get v = √(2gh). Plugging in the values, we get v = √(2)(9.8 m/s²)(6,378,000 m) = 7,905 m/s.

4. How long does it take for the object to reach the surface from 400 km above?

Using the equation d = 1/2at^2, where d is the distance traveled, a is the acceleration (9.8 m/s²), and t is the time taken, we can solve for t. Plugging in the values, we get t = √(2d/a) = √(2)(400,000 m)/(9.8 m/s²) = 2,020 seconds or 33.67 minutes.

5. What is the potential energy of the object when it reaches the surface of the earth?

Since the object is now at the surface, its height above the surface is zero. Therefore, the potential energy would also be zero, as potential energy is dependent on the height above the surface.

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