Orbit around the Earth

1. Apr 20, 2016

Tissue

1. The problem statement, all variables and given/known data
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

2. Relevant equations
F=GMm/r2 F=mv2/r

3. The attempt at a solution
I tried to use the formula above which is GM/r=v2, however I substituted both 6.81x106 ,7.53x106 and took the average of two of them , the answer still did not match with the MC answer which is 7840m/s
I am curious about what should be the radius, and actually the orbit is not a perfect circle with a perigee and an apogee , is it acceptable to use the equation in circular motion . Thank you.

2. Apr 20, 2016

PeroK

Note that the centripetal acceleration is related to the curvature, so you can't use the equation for a circular orbit.

What do you know about angular momentum?

3. Apr 20, 2016

Tissue

The angular momentum is conserved as the gravitational pull by the Earth is always perpendicular to the motion, no matter where is the position of the Sputnik I . And L=mvr

4. Apr 20, 2016

PeroK

Yes. And total energy (kinetic & potential) is conserved also. Can you do something with that?

To be precise, the angular momentum is $L = mv_ar$ where $v_a$ is the angular component of velocity.

And, $v_a = v$ only at the turning points, where the radial component of the velocity is 0.

Last edited: Apr 20, 2016
5. Apr 20, 2016

Tissue

OK Thanks a lot.
As angular momentum is conserved. mv(for apogee)*7.53*106=mv(for perigee)*6.81*106
Then v(for apogee)=0.90428v(for perigee)
M is equal to the mass of earth
As energy is also conserved. mv(for perigee)/2-GMm/7.53*106=mv(for apogee)/2-GMm/6.81*106

6. Apr 20, 2016

PeroK

Now you've reached this level of physics, you ought to consider working more algebraically. I would say:

$mv_1r_1 = mv_2r_2$ where $r_1$ is the perigee etc.

Numbers like $6.81*10^6$ are just going to clutter up your equations and hide the physics.

Solve for $v_1$ in terms of $r_1, r_2$ then plug the numbers in at the end.

7. Apr 20, 2016

ehild

The v-s have to be squared in the kinetic energies.

8. Apr 21, 2016

Tissue

Thank you very much. Get the answer finally.
May I ask how to quote the formula for the angular momentum in a more systematic way like you?