Orbit around the Earth

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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.
 

Answers and Replies

  • #2
PeroK
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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?
 
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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
 
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PeroK
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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.
 
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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
PeroK
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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.
 
  • #7
ehild
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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.
 
  • #8
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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?
 

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