Question

How fast does a satellite have to leave Earth's surface to reach an orbit with an altitude of 895 km?

please help. I got 3922.05 m/s but I am not sure.

Answer 1

Can you show me your work?

Answer 2

that is not the right answer!

Answer 3

Remember that you must measure the radius from the centre of the earth, and that the gravitational acceleration is not g. It is GM/r^2.

Centripetal force = mv^2/r
This must equal gravitational force
(m cancels out)

Answer 4

I used Ek=deltaEp

Ek=Ep2-Ep1

1/2mv^2=(-GMm/r2)-(-GMm/r1)

I solved for v and got that number.

M is the earth's mass, r2 is the distance from earth's centre to required altitude (7275000 m), and r1 is the radius of earth.

Clearly, I am doing something wrong here. I didn't set up my equation correctly.

Answer 5

@Mashy

Answer 6

and why is the total potential energy 0 when you escape earth's gravitational pull?

Answer 7

Please helppp!!

Answer 8

Of course PE becomes 0. The entire energy is now kinetic.

Answer 9

Maybe the problem is with the phrasing of the question.

If the question was (ignoring air resistance) how fast must a projectile be fired vertically upwards to reach a height of 895 km then you can address it with the KE=PE approach.

Once it has reached this height it would of course fall straight back down again (and much of its KE would be lost to air friction in real situation - so initial KE would need to be much higher)

However - to achieve orbit it must have an orbital velocity, and therefore its KE is not zero at the given height.

The centripetal force for a circular orbit is mv^2/r
The gravitational force is GMm/r^2

SO I believe the question should be phrased "What velocity must a satellite have to remain in orbit at 895km"

Answer 10

OK - I have found your question on another site.
Your answer of 3922 is correct for 'reaching the altitude of 895km' as I proposed.

That was a separate question on the same paper.

1. How fast must a rocket leave the Earth’s surface to reach a maximum height of 895 km above the surface of the Earth (assume the rocket is simply going straight up)? (3920 m/s)

2. How fast must a rocket leave the Earth’s surface in order to get into an orbit at an altitude of 895 km above the surface of the Earth? (8380 m/s)

Answer 11

satecphysics12s.pbworks.com/f/Gravitational+Energy+Practice.doc‎

Answer 12

Yes.. you see.. what you have calculated only allows it to reach that altitude

so with that speed.. when it reaches that altitude.. its k.e gonna become zero.. and gravity will make the rocket its biaach.. and the rocket wil fall back

but in order for the satellite to ORBIT (it needs to have certain k.e)..

Answer 13

@coolsday Over to you.......

Answer 14

What certain Ek would that be?

Answer 15

well refer your orbiting satellite energy ..

any orbiting satellite must have a total energy of
\[E_{orbit} = -\frac{GMm}{2r}\]

Answer 16

to orbit at a distance of r from the centre!

Answer 17

Isn't that the energy that is needed for a satellite to escape from its current orbit?

Answer 18

No o.o

Answer 19

? In my textbook, it says, the additional energy it needs to escape from Earth's gravitational pull is GMm/2r

Answer 20

Its for this reason.. you have to derive it.. cause once you derive it.. everything becomes clear :P..


and yes.. since it already have - GMm/2r (when orbiting at a distance r).. it needs +GMm/2r energy to make it zero.. or just reach infinity

Answer 21

I'd also like to say that the two questions side by side are very misleading.
It seems ot suggest that simply by increasing the velocity then an orbit will be obtained.

A rocket fired VERTICALLY up will never go into orbit. It will either escape earths gravity entirely, or fall back to earth.

It is important to understand that the velocity required for orbit is tangential to the orbit - not perpendicular to earth. Unless it achieves this tangential velocity it will not orbit.

The question is not well phrased in my opinion.

Answer 22

You said that any orbiting satellite must have a total energy of
Eorbit=−GMm/2r, why is this the case? @Mashy

Then in what direction must the rocket be fired to go into orbit? @MrNood

Answer 23

"s important to understand that the velocity required for orbit is tangential to the orbit - not perpendicular to earth. Unless it achieves this tangential velocity it will not orbit."

that is wrong

in reality.. you fire at any angle it should orbit .. (or try to orbit.. cause if planet is too big and comes in the way.. it may crash)

it ll usually orbit in elliptical shape.. the orbital velocity is only if you want it to orbit in a perfect circle.!!

Answer 24

@coolsday
you can derive that

at distance r

the potential energy = -GMm/r

and it needs to orbit with orbital speed .. therefore.. if you find the required k.E
ti turns out to be +GMm/2r

so when you add them up, total energy = -GMm/2r

Answer 25

"s important to understand that the velocity required for orbit is tangential to the orbit - not perpendicular to earth. Unless it achieves this tangential velocity it will not orbit."

that is wrong



@ mashy - I don't think it IS wrong Velocity is always tangential to orbit - it may not be tangential to the earth if the orbit is elliptical.

The direction of rocket propulsion progressively changes from vertical to tangential.

The orbital velocity is constant if circular, otherwise it follows the 'equal swept area' rule for an elliptical orbit.

Answer 26

Ofcourse its tangential to the ORBIT.. :P..
i was just saying.. that a rocket can orbit with any speed and can be projected at ANY ANGLE..

when you project the rocket at any angle, the orbit becomes such that, at that point.. that angle would be tangential :P..

Answer 27

@mashy What I said was a rocket fired VERTICALLY

That will not enter orbit and will either escape (if its speed is sufficient) or fqall back to earth.

That is the exception to your statement

Answer 28

yea.. but if you think of the planet as a POINT MASS ... then its not xD.. m speaking theoretical here..

in that case, it would swing back and forth.. ofcourse its not called as an orbit.. but then again straight line IS a degenerated ellipse.. !

Answer 29

Ok, guys. Thanks for all your help! Can you help me on another one? @Mashy

Answer 30

A satellite is in orbit around Earth at an altitude of 3700 km. How much speed must it increase, in km/h, to escape Earth's force of gravity?

I got 9381.11 km/h.