# Velocity of satellite at specific altitude

• Biosyn
In summary: I missing something?Oh, right. Well, the answer says it takes the satellite 6 hours to complete one orbit around Earth which is about 3 hours shorter than what I came up with (9 hours). Am I... am I missing something?Your answer is correct. The period for a satellite at an altitude of 10,000 miles should be approximately 9 hours and 20 minutes. The answer given in the conversation is incorrect.
Biosyn

## Homework Statement

How fast must a satellite 10,000 miles above Earth's surface travel, and how long does it take to complete one orbit of Earth?

## Homework Equations

$v^2 = g*r$

## The Attempt at a Solution

10,000 miles + radius of Earth = 14,000mi =

$v^2 = (9.8)(2.253*10^7m)$

v = 14859.13 m/s

Is there something I'm doing wrong or overlooking?

Is gravitational acceleration constant?

jhae2.718 said:
Is gravitational acceleration constant?

$\frac{g_1}{g_2}$ = ($\frac{r_2}{r_1}$)^2

$\frac{22}{g_2}$ = ($\frac{14000mi}{4000mi}$)^2

$g_2$ = 1.796mi/h/s

$v^2$ = (1.796)(14000)

v = 158.56 mi/h ?

Check your units. Also, what units are your value for ##g_1##? Recall ##g_0 \approx 32.17\text{ ft/s}^2##.

jhae2.718 said:
Check your units. Also, what units are your value for ##g_1##? Recall ##g_0 \approx 32.17\text{ ft/s}^2##.

$g_2 = 1.7959 mph$

(1.795mph)*(14000mi) = v^2

v = 158.56 mi/√hEdit: For $g_1$ I used 22mph because the other units are in miles.

Biosyn said:
$g_2 = 1.7959 mph$

(1.795mph)*(14000mi) = v^2

v = 158.56 mi/√h

Edit: For $g_1$ I used 22mph because the other units are in miles.

Sorry to say it, but this isn't making any sense. The quantity g is supposed to be an acceleration, but you have units of speed instead.

My approach to solving this problem would be to look at Newton's Universal Law of Gravitation. I would also ask myself: considering that the object is moving in a circle at a constant speed, what kind of force must be present?

cepheid said:
Sorry to say it, but this isn't making any sense. The quantity g is supposed to be an acceleration, but you have units of speed instead.

My approach to solving this problem would be to look at Newton's Universal Law of Gravitation. I would also ask myself: considering that the object is moving in a circle at a constant speed, what kind of force must be present?

Okay sorry, I meant 22 mi/h/s...

and g_2 = 1.7959 mi/h/s at 10,000 miles above Earth's surface.

Check what you're doing with ##g_2##...the unit you should have is mi/(h*s)...so, what conversion do you need to do to your result?

You should maintain units throughout your calculations to avoid these types of errors.

Also, you should carry more precision for ##g_1## and ##R_\oplus##.

Okay, I'm getting confused. In the beginning I converted everything to meters.

altitude = 1.609*10^7m + 6.3675*10^6m
gravitational acceleration = 9.8m/s^2
v = ?

v^2 = g*r
v = sqrt(g*r)

v = sqrt( 9.8 * (1.609*10^7m + 6.3675*10^6m))
v = 14835.2m/s

But this doesn't equal 9400 mph.

I was using this website for help: http://www.ux1.eiu.edu/~cfadd/1150/05UCMGrav/Sat.html

Biosyn said:
Okay, I'm getting confused. In the beginning I converted everything to meters.

altitude = 1.609*10^7m + 6.3675*10^6m
gravitational acceleration = 9.8m/s^2
v = ?

v^2 = g*r
v = sqrt(g*r)

v = sqrt( 9.8 * (1.609*10^7m + 6.3675*10^6m))
v = 14835.2m/s

But this doesn't equal 9400 mph.

I was using this website for help: http://www.ux1.eiu.edu/~cfadd/1150/05UCMGrav/Sat.html

Suddenly we're back to square one. It has already been explained that you're using the wrong value for g, the one that applies at the Earth's surface, not 10,000 miles up. EDIT: This is why I like my approach of deriving the equation for the speed from Newton's Law of Gravitation. It's conceptually clearer, because stuff isn't hidden in "g."

Biosyn said:
gravitational acceleration = 9.8m/s^2

##g \not = 9.8\text{ m/s}^2## that far from Earth. Recall that ##F_g \propto \frac{1}{r^2}##.

Use Newton's gravitational law to find ##g## at the altitude specified in the problem.

(Generally, SI units are almost exclusively used in the space world.)

jhae2.718 said:
##g \not = 9.8\text{ m/s}^2## that far from Earth. Recall that ##F_g \propto \frac{1}{r^2}##.

Use Newton's gravitational law to find ##g## at the altitude specified in the problem.

(Generally, SI units are almost exclusively used in the space world.)

Okay, $g_2$ = (6.6742*10^(-11))*((5.9736*10^24)/(16371011.61))

$g_2$ = 1.4875 m*s^-2

Biosyn said:
Okay, $g_2$ = (6.6742*10^(-11))*((5.9736*10^24)/(16371011.61))

$g_2$ = 1.4875 m*s^-2

You're using the wrong radius. ##R = R_\oplus + r##, where ##R## is the total radius, ##R_\oplus## is the radius of Earth, and ##r## is the geometric altitude of the satellite.

Your value is off by an order of magnitude.

My advice would be to solve the problem in terms of symbolic variables until you have an expression for ##v## in terms of the known variables, and only then substitute numerical values.

jhae2.718 said:
You're using the wrong radius. ##R = R_\oplus + r##, where ##R## is the total radius, ##R_\oplus## is the radius of Earth, and ##r## is the geometric altitude of the satellite.

Okay.

10,000 miles = 1.609*10^7 meters

R = (6.37101*10^6 + 1.609*10^7)
R= (22461010)
R^2 = 5.044*10^14m

g_2 = (6.6742*10^-11)*(5.044*10^14)
g_2 = 0.790424

v^2 = 0.790424 * 22461010
v = 4213.51 m/s

Edit: Which is approximately 9400 mph. Thanks for helping me! :)

No problem!

jhae2.718 said:
No problem!

Actually there's one more question I have which is finding the time it takes the satellite to orbit the Earth once.

T = 2*pi*sqrt((r^3)/(G*m))

T = 2*pi* Sqrt($\frac{22461010^3}{G*m_E}$)T = 33627.87 seconds
T = 9 hours 20 minutes

Biosyn said:
Actually there's one more question I have which is finding the time it takes the satellite to orbit the Earth once.

T = 2*pi*sqrt((r^3)/(G*m))

T = 2*pi* Sqrt($\frac{22461010^3}{G*m_E}$)

T = 33627.87 seconds
T = 9 hours 20 minutes

This is fine, but there's an easier way to calculate it, which is to say speed = distance/time, where the distance is the circumference of the orbit.

Since you *already* calculated the speed, you can just plug in into this equation. If you use the equation that has r , G, and M_E in it, then you are implicitly calculating the speed *again*, which is a waste of time, since you already know it.

cepheid said:
This is fine, but there's an easier way to calculate it, which is to say speed = distance/time, where the distance is the circumference of the orbit.

Since you *already* calculated the speed, you can just plug in into this equation. If you use the equation that has r , G, and M_E in it, then you are implicitly calculating the speed *again*, which is a waste of time, since you already know it.

Oh, right.
Well, the answer says it takes the satellite 6 hours to complete one orbit around Earth which is about 3 hours shorter than what I came up with (9 hours). Am I wrong?

Biosyn said:
Oh, right.
Well, the answer says it takes the satellite 6 hours to complete one orbit around Earth which is about 3 hours shorter than what I came up with (9 hours). Am I wrong?

No, you just used the wrong radius for the orbit. The orbit is a circle. The radius of a circle is the distance between the *centre* of that circle and each point on the circle. In this case, what is the radius of the circle? (Hint: it is not 10,000 miles).

cepheid said:
No, you just used the wrong radius for the orbit. The orbit is a circle. The radius of a circle is the distance between the *centre* of that circle and each point on the circle. In this case, what is the radius of the circle? (Hint: it is not 10,000 miles).

It's the radius of Earth + the distance of the satellite from the surface of Earth. This is what I used though...

Edit: I used 22461010 meters.

6.3675*10^6 + 1.609*10^7 = 22457500 meters

^Using this radius, I still get around 9 hours.

Biosyn said:
It's the radius of Earth + the distance of the satellite from the surface of Earth. This is what I used though...

Edit: I used 22461010 meters.

6.3675*10^6 + 1.609*10^7 = 22457500 meters

^Using this radius, I still get around 9 hours.

Sorry, I misread what you said before. I thought that *you* got 6 hours and the answer key said 9.

You got ~9 hrs using two independent methods. So that's consistent. And (2*pi*14,000 mi)/(9400 mph) is definitely about nine and a third hours. (without a calculator, 2pi is ~6, and 6*14 is 84, and 84/9 is just above 9).

If the answer key says six, then *they* made the mistake with the orbit radius that I referred to before.

cepheid said:
Sorry, I misread what you said before. I thought that *you* got 6 hours and the answer key said 9.

You got ~9 hrs using two independent methods. So that's consistent. And (14,000 mi)/(9400 mph) is definitely about nine and a third hours. If the answer key says six, then *they* made the mistake with the orbit radius that I referred to before.

Oh, okay. Thank you for the help! :)

Approx. 9 hours is correct. If you work backwards, the provided answer uses ##R = R_\oplus + 5000\text{ mi}## instead of ##R = R_\oplus + 10000\text{ mi}##.

## 1. What is the formula for calculating the velocity of a satellite at a specific altitude?

The formula for calculating the velocity of a satellite at a specific altitude is V = √(G*M/R), where V is the velocity, G is the gravitational constant, M is the mass of the Earth, and R is the distance from the center of the Earth to the satellite.

## 2. How does the velocity of a satellite change with increasing altitude?

As the altitude of a satellite increases, the velocity decreases. This is because the gravitational force between the satellite and the Earth weakens as the distance between them increases, resulting in a slower orbital speed.

## 3. What is the relationship between the velocity of a satellite and its orbital period?

The velocity of a satellite is directly proportional to its orbital period. This means that as the velocity increases, the orbital period also increases. This relationship is described by Kepler's third law of planetary motion.

## 4. How does the velocity of a satellite at a specific altitude affect its stability?

The velocity of a satellite at a specific altitude is directly related to its stability. If the satellite is moving too slowly, it may fall towards the Earth due to the strong gravitational pull. On the other hand, if the satellite is moving too quickly, it may fly away from the Earth and be unable to maintain its orbit.

## 5. How is the velocity of a satellite at a specific altitude measured?

The velocity of a satellite at a specific altitude can be measured using various methods such as radar, laser ranging, and Doppler shift measurements. These techniques involve sending out signals to the satellite and measuring the time it takes for the signals to return, which can then be used to calculate the satellite's velocity.

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