# Orbit around the Earth

## 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

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.

Related Introductory Physics Homework Help News on Phys.org
PeroK
Homework Helper
Gold Member
2020 Award
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?

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
Homework Helper
Gold Member
2020 Award
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.

Last edited:
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
Homework Helper
Gold Member
2020 Award
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
Homework Helper
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.

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?