# What Is the Correct Calculation for Satellite Speed in a Stable Orbit?

• BitterSuites
In summary, the satellite must have a centripetal acceleration inversely proportional to the square of the radius of its orbit in order to stay in a stable orbit. The speed of the satellite is 9.0 m/s.
BitterSuites
[SOLVED] Centripetal Acceleration

## Homework Statement

In order for a satellite to move in a stable circular orbit of radius 6689 km at a constant speed, its centripetal acceleration must be inversely proportional to the square of the radius of the orbit.

What is the speed of the satellite? The universal gravitational constant is 6.67259e-11 and the mass of the Earth is 5.98e24.

a = v^2/r
F = Gm1m2/r^2
F = ma

## The Attempt at a Solution

Ok, a = v^2/r becomes v = sqrt of ar.

From the problem, a = r^-1/2 (or am I wrong?) so a = .012227

sqrt of ar = sqrt of 81.9863
v=9.0-4358

Centripetal acceleration=force of gravity

(m2)(v^2/r)=(G*m1*m2)/r^2

You can find acceleration from there.

I already have acceleration from r=-1/2 as told from the problem itself. I need to solve for v. Can I use the above equation if I don't know v or m1?

There is only one force acting on the satellite, the force of gravity. You know that $$F_g=\frac{GMm}{r^2}$$.

By Newton's 2nd you also know that $$\sum F=ma$$ .

Now ask yourself how you can use this along with what you were given to find v.

Last edited:
BitterSuites said:
From the problem, a = r^-1/2 (or am I wrong?) so a = .012227

Yes, you're right - you're wrong!

The question is telling you that a is proportional to r^-1/2. All it means is that a = F/m, and F is proportional to r^-1/2, so a must be also.

This is a questioner intending to be helpful, but actually being a little confusing.

Your "relevant equations" are correct, so just solve them (and ignore the help).

mv^2/r =GMm/r^2
v^2=GM/r^2
v= sqrt GM/r^2

GM=gRe^2 as well

Ok, so I did it this way and still ended up with the incorrect answer. Where am I still messing up?

F=Gm1m1/r^2
9.8 = 6.67259e-11*M*5.98e24/6689^2
4.38471e8=6.67259e-11*M*5.98e24
7.33242e-17=6.67259e-11*M
M=.000001

F=ma
9.8=.000001a
a = 8.91812e6

a=v^2/r
8.91812e6 = v^2/6689
v^2=5.96533e10
v=244240

So where did I go wrong again? :D

BitterSuites said:
Ok, so I did it this way and still ended up with the incorrect answer. Where am I still messing up?

F=Gm1m1/r^2
9.8 = 6.67259e-11*M*5.98e24/6689^2
4.38471e8=6.67259e-11*M*5.98e24
7.33242e-17=6.67259e-11*M
M=.000001

F=ma
9.8=.000001a
a = 8.91812e6

a=v^2/r
8.91812e6 = v^2/6689
v^2=5.96533e10
v=244240

So where did I go wrong again? :D
it's 6689km... not 6689m

I tried v = sqrt GM/r^2 and it is also incorrect.

mysqlpress said:
it's 6689km... not 6689m

Wow. I'm special. Let me throw that in and see what happens :D

BitterSuites said:
I tried v = sqrt GM/r^2 and it is also incorrect.
it's my typo
should be v=sqrt GM/R

you see, from my deviation :)

Phew. On the last try I got it correct. My calculations above worked once I converted km to m. *judges himself harshly*

v = sqrt GM/r^2 didn't work even with the km converted to m.

Ah :) Thanks for everyone's help! I'm marking it solved now.

## 1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and is responsible for keeping the object in its circular motion.

## 2. How is centripetal acceleration calculated?

The formula for calculating centripetal acceleration is a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

## 3. What is the difference between centripetal acceleration and centrifugal force?

Centripetal acceleration is a real force that keeps an object in its circular motion, whereas centrifugal force is a perceived force that appears to push an object outward from the center of the circle. In reality, centrifugal force is simply the result of inertia.

## 4. How does centripetal acceleration relate to Newton's laws of motion?

Centripetal acceleration is a result of Newton's first law, which states that an object will remain in its state of motion unless acted upon by an external force. In the case of circular motion, the centripetal acceleration provides the necessary force to keep the object moving in its circular path.

## 5. What are some real-life examples of centripetal acceleration?

Some common examples of centripetal acceleration include the motion of a car around a curve, the rotation of a spinning top, the orbit of planets around the sun, and the motion of a roller coaster around a loop. Any object moving in a circular path experiences centripetal acceleration.

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