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

  • Thread starter Thread starter Scarlett
  • Start date Start date
  • Tags Tags
    Rest Surface
AI Thread Summary
A 1600 kg object falls from 400 km to 100 km above Earth's surface, generating 8.3x10^8 J of heat energy. The calculations involve gravitational potential energy and kinetic energy equations, leading to the need for accurate values of radius in meters. A mistake was identified in the calculation of the difference in gravitational potential, specifically in converting kilometers to meters. The correct difference should yield a value of 6x10^-9 instead of 7.37 x 10^-12. The discussion highlights the importance of unit conversion in physics problems.
Scarlett
Messages
3
Reaction score
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:
Physics news on Phys.org
Your solution is correct, but the mistake lies in 1/r2 - 1/r1, this equals 6x10^-9, not 7.37 x 10^-12.
 
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
 
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 :)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top