# Acceleration due to gravity

## Homework Statement

Sputnik I was launched into orbit around Earth in 1957. It had a perigee (the closest approach to
Earth, measured from Earth's center) of 6.81 × 106 m and an apogee (the furthest point from
Earth's center) of 7.53 × 106 m. What was its speed when it was at its perigee? The mass of Earth
is 5.97 × 1024 kg and G = 6.67 × 10- 11 N ∙ m2/kg2

F=GMm/r2 F=mv2/r

## The Attempt at a Solution

As energy is also conserved. mv(for perigee)/2-GMm/7.53*106=mv(for apogee)/2-GMm/6.81*106
But how can I find v(for apogee)? Thus v(for perigee)?
I can't find the MC answer which is 7840m/s

http://www.enotes.com/homework-help/sputnik-was-launched-into-orbit-around-earth-1957-566404

Thanks

## Answers and Replies

gneill
Mentor
There's another conservation law that will help here.

Angular momentum? L=rp=rmv
But still i need to find v for apogee... ehich i dont know how to get the answer avoiding this unknown variable...

gneill
Mentor
Yes, angular momentum is conserved over the whole orbit. What does that tell you about the angular momentum at both apogee and perigee? Can you express that as an equation?

Janus
Staff Emeritus
Gold Member
Can you determine the period of the orbit from what you are given? And if so, what does that give you?

Yes, angular momentum is conserved over the whole orbit. What does that tell you about the angular momentum at both apogee and perigee? Can you express that as an equation?

ra x va = rp x vp
Right? Without vp, how to find va?

Can you determine the period of the orbit from what you are given? And if so, what does that give you?
T=2(pi)/(angular velosity) = 2(pi)r/(average v)?

gneill
Mentor
ra x va = rp x vp
Right? Without vp, how to find va?
This gives you one relationship between va and vp. Energy conservation gives you another. Two equations, two unknowns.

T=2(pi)/(angular velosity) = 2(pi)r/v?
Do you agree with this?
What is r and what is v?
Thanks

gneill
Mentor
Do you agree with this?
What is r and what is v?
Thanks
If the orbit were circular then I'd agree. But the velocity and radius both vary over time for an elliptical orbit.

Use the conservation laws: energy and angular momentum.

Janus
Staff Emeritus
Gold Member
Do you agree with this?
What is r and what is v?
Thanks
Not quite what I was looking for. There is a fairly direct way to get the answer to the problem, but it assumes some knowledge of the nature of orbits. I was trying to gauge just how much you knew about orbits in order to know in what direction to lead you in.

Use the conservation laws: energy and angular momentum.
Angular momentum: Rmva=rmvp
Energy: 0.5ka^2 ; 0.5mv^2; -gMm/r => how should i arrange them?
Thanks

I have only little time left before my rxam is coming... hope you can tell me the sooner the better for me to try and understand them... thank you very much.

gneill
Mentor
Angular momentum: Rmva=rmvp
Okay. Note that the m's can cancel.
Energy: 0.5ka^2 ; 0.5mv^2; -gMm/r => how should i arrange them?
Apply energy conservation for the two orbit locations.

I know i should use energy conservation.... but how to arrange them? If 0.5k(ap)^2+0.5m(vp)^2+gMm/r=0.5K(aa)^2+0.5m(va)^2+GMm/r
How come there are so many variable?! Am i having something wrong?
So how can i arrange them???
Thanks...

(Only little time left from test...)

gneill
Mentor
I know i should use energy conservation.... but how to arrange them? If 0.5k(ap)^2+0.5m(vp)^2+gMm/r=0.5K(aa)^2+0.5m(va)^2+GMm/r
How come there are so many variable?! Am i having something wrong?
I don't know what your "k" terms are meant to represent. What type of energy do they correspond to? What is k?

You need to get your signs right for the gravitational potential energy. Gravitational potential energy in this context takes its reference point at infinite distance, so its sign is always negative.

You have two given locations, apogee and perigee, for which you have separate orbit radii. Be sure to identify them in your equation with separate variable names.

K is spring constant, as such the planet is oscillating like a spring
It is correspond to elastic potential energy

gneill
Mentor
K is spring constant, as such the planet is oscillating like a spring
It is correspond to elastic potential energy
There's no spring here. The "spring" action of the orbit is accounted for by the trading of gravitational potential energy and kinetic energy.

haruspex
Homework Helper
Gold Member
2020 Award
K is spring constant, as such the planet is oscillating like a spring
It is correspond to elastic potential energy
To the extent that the satellite oscillates in its orbit, it is a result of the gravitational attraction term you already have in the equation, not from some mysterious piece of elastic tethering it to Baikonur.
Besides, gravitation follows an inverse square law of attraction, while elastic follows a linear law.

ChloeYip
To the extent that the satellite oscillates in its orbit, it is a result of the gravitational attraction term you already have in the equation, not from some mysterious piece of elastic tethering it to Baikonur.
Besides, gravitation follows an inverse square law of attraction, while elastic follows a linear law.

Thank you very much.
I can find the answer now.