Can someone Check this (Energy Question)

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The discussion centers on calculating the total energy of a geosynchronous satellite with a mass of 1500 kg at an altitude of 325 km and an orbital speed of 5000 m/s. The initial calculations for gravitational potential energy (Ep) and kinetic energy (Ek) were presented, resulting in a total energy of approximately 2.353x10^10 joules. However, it was noted that using the formula Ep = mgh is only accurate for objects near Earth's surface, and the gravitational potential energy at higher altitudes should be calculated using the formula Ep = -Gm1m2/r. Participants emphasized the importance of using the correct gravitational potential energy equation for accurate results. The conversation highlights the need for precision in physics calculations, especially at varying altitudes.
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Q: Calculate the total engery of a geosynchonous satellite (one that orbits over a fixed spot) with amass of 1500 kg, orbiting Earth at height 325 km (325 000m) with an orbital speed of 5000 m/s or 5.0x10^3

My Solution: Et = Ep + Ek
Ep = Mg delta H
Ek = 1/2 (m) V^2
thus Et = Mg(H) + 1/2 (m) v^2
(1500)(9.8)(325000) + 1/2(1500) (5000)^2
= 4.78x10^9 + 1.875x10^10
=2.353x10^10 j or 2.353x10^7 Kj

Is there a mistake in my calculations or has this question been done correctly?
 
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quickslant said:
Q: Calculate the total engery of a geosynchonous satellite (one that orbits over a fixed spot) with amass of 1500 kg, orbiting Earth at height 325 km (325 000m) with an orbital speed of 5000 m/s or 5.0x10^3

My Solution: Et = Ep + Ek
Ep = Mg delta H
Ek = 1/2 (m) V^2
thus Et = Mg(H) + 1/2 (m) v^2
(1500)(9.8)(325000) + 1/2(1500) (5000)^2
= 4.78x10^9 + 1.875x10^10
=2.353x10^10 j or 2.353x10^7 Kj

Is there a mistake in my calculations or has this question been done correctly?
using g = 9.8m/s/s is valid only near earth. What can you say about GPE at those heights?
 
i have no idea? is it 0? or is it very small? i have no idea
 
Look up the general expression for the gravitational PE between two masses (such as the Earth and a satellite).

As PhanthomJay pointed out, mgh is only useful for finding changes in PE for objects close to the surface of the earth.
 
ok.. Ep = Gm1m2/r
 
quickslant said:
ok.. Ep = Gm1m2/r
Don't forget the minus sign: Ep = -Gm1m2/r
 
ok thank you
 
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