What is the speed of Sputnik I when it was at its perigee around Earth in 1957?

[Tissue]

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

Homework Equations

F=GMm/r² F=mv²/r

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.

 

[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?

 

[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

 

[PeroK]

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

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.

 

[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

 

[PeroK]

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

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.

 

[ehild]

As energy is also conserved. mv(for perigee)/2-GMm/7.53*10⁶=mv(for apogee)/2-GMm/6.81*10⁶

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

 

[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?