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

• Tissue
In summary, 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.
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

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.

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

Tissue said:
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

Tissue said:
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.

Tissue said:
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?

## What is the orbit around the Earth?

The orbit around the Earth refers to the path that an object takes as it revolves around the Earth due to the Earth's gravitational pull.

## What is the shape of the orbit around the Earth?

The shape of the orbit around the Earth is elliptical, meaning it is oval-shaped. This is due to the combined gravitational forces of the Earth and the Sun.

## What factors affect the orbit around the Earth?

The orbit around the Earth is affected by several factors, including the object's mass, speed, and distance from the Earth, as well as the gravitational pull of other celestial bodies.

## What is the difference between a geostationary and a geosynchronous orbit around the Earth?

A geostationary orbit around the Earth is when the object stays in a fixed position above a specific point on the Earth's equator, while a geosynchronous orbit is when the object's position above the Earth repeats at the same time each day.

## How is the orbit around the Earth calculated?

The orbit around the Earth is calculated using Newton's laws of motion and the law of universal gravitation, which takes into account the masses and distances of the objects involved.

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