Why Is Achieving Orbit Considered Halfway to Anywhere?

AI Thread Summary
Achieving orbit is often viewed as halfway to anywhere because the energy required to reach low Earth orbit is significantly less than that needed to escape Earth's gravitational pull entirely. The total energy equation combines kinetic and gravitational potential energy, illustrating that escape velocity is independent of the satellite's mass. By setting the total energy to zero, the mass cancels out, allowing for the calculation of velocity without needing the satellite's mass. The escape velocity is approximately 7 miles per second, highlighting the efficiency of reaching orbit compared to complete escape. This understanding underscores the importance of orbital mechanics in space travel.
dswatson
Messages
36
Reaction score
0
The science fiction writer Robert Heinlein once said, "If you can get into orbit, then you're halfway to anywhere". Justify this statement by comparing the minimum energy needed to place a satellite into low Earth orbit (h=400km) to that needed to set it completely free from the bonds of Earth's gravity. Neglect any effects due to air resistance.

E = KE + U
E = 1/2mv^2 + U
U = -(int)[F*dr]
U = -G(Mm/r^2)
E = m( (1/2)v^2 - G(M/r^2)

m = mass satellite
M = mass earth

Im stuck because I am not given a mass for the satellite and I know that excape velocity does not depend on mass. It is approxmately 7mi/s but I am unsure of how to show that energy is not mass dependent and then solving for the equation with no mass.
 
Physics news on Phys.org
Well, first of all, the total energy is

E = m\left( \frac{v^2}{2} - \frac{GM}{r}\right)

(you got the power wrong). It depends on m, but if you set it equal to zero E=0, then m cancels and you can solve for v without knowing m.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top