How much work is done when a satellite is launched into orbit?

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Homework Help Overview

The discussion revolves around calculating the work done to launch a satellite into a geostationary orbit, specifically focusing on the gravitational work required to move the satellite from the Earth's surface to an altitude of 35,800 km. The problem involves concepts from gravitational physics and energy calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to compute the work done using Newton's law of universal gravitation and expresses confusion regarding a discrepancy between their calculation and the textbook answer. Other participants question whether potential energy is the only factor in achieving geostationary orbit, leading to discussions about the necessity of kinetic energy as well.

Discussion Status

Participants are exploring the relationship between potential and kinetic energy in the context of orbital mechanics. Some guidance has been offered regarding the need to consider both forms of energy, and there is an ongoing inquiry into deriving kinetic energy from Newton's second law.

Contextual Notes

There is a noted discrepancy between the original poster's calculated work and the textbook's suggested answer, prompting further investigation into the assumptions and calculations involved. The discussion also highlights the need for additional information regarding the velocity required for maintaining orbit.

tuki
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Homework Statement


A Satellite is brought up into a geostationary orbit (altitude 35800km measured from the surface of the earth). Satellite weights 1000.0kg. How much work is required to bring satellite from a surface of the Earth to
geostationary orbit?

Homework Equations


Newton's law of universal gravitation
$$ F = G \frac{m_1m_2}{r^2} $$
Work is defined as:
$$ W \int F \cdot s ds $$

The Attempt at a Solution


I want to compute work done by Newton's law of universal gravitation when
moving satellite from the surface of the Earth to the orbit. It should be done with:h0 is surface of the Earth (about 6371 km)
h1 is geostationary orbit (about 6371 km + 35800 km = 42171km)
m1 is mass of the satellite (about 1000 kg)
m2 is mass of the Earth (about 5.972E24)

$$ W = \int_{h_0}^{h_1} G \frac{m_1 m_2}{r^2} dr = - G \frac{m_1 m_2}{h_1} - (- G \frac{m_1 m_2}{h_0} $$
$$ = G m_1 m_2 (\frac{1}{h_1} - \frac{1}{h_0}) \approx 5.3108\cdot 10^{10} \text{ J} $$

However, our textbook suggests that the correct answer would be 5.77E10 Joules. I can't exactly spot out where the mistake is.
 
Physics news on Phys.org
Is giving the satellite potential energy the only thing you need to do to put it in geostationary orbit?
 
Orodruin said:
Is giving the satellite potential energy the only thing you need to do to put it in geostationary orbit?
Yes you would need the velocity too in order to stay in orbit.
 
Can you see that that's your answer, or is still unclear?
 
Bandersnatch said:
Can you see that that's your answer, or is still unclear?
Yes, I can get the correct answer by adding kinetic energy from the speed $$ E_{kinetic} = \frac{1}{2} mv^2 $$ to the potential energy I already computed. I wonder if it would be possible to derive the work required to accelerate an object to a given speed (when mass is known) starting from Newton's second law? $$ F = ma $$ I mean you start from Newton's second law and end up with a formula for kinetic energy?
 
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tuki said:
I wonder if it would be possible to derive the work required to accelerate an object to a given speed (when mass is known) starting from Newton's second law?
F=maF=ma​
F = ma I mean you start from Newton's second law and end up with a formula for kinetic energy?
Why don’t you try and see what you get?
 

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