Calculate Satellite Speed & Period at 400 km Height

[max0005]

Homework Statement

Estimate the speed of a satellite orbiting Earth at an height of 400km. Also determine the period of the orbit.

Homework Equations

Centripetal acceleration: $$F=\frac{mv^2}{r}$$
Newton's Law of Universal gravitation: $$F=\frac{G(MassEarth)(MassSatellite)}{R^2}$$
G = 6.67*10^-11
Mass of Earth: 6*10^24kg

The Attempt at a Solution

I now have that:
$$\frac{(MassSatellite)v^2}{400} = \frac{G(MassEarth)(MassSatellite)}{1.6*10^5}$$

Therefore:

$$v=\sqrt\frac{G(MassEarth)}{400}$$

Where G has a fixed value of 6.67*10^-11.

I therefore get that v has a value of 2*10^6 km/h... Isn't this excessive? This would mean that if we take the radius of Earth and estimate it to about 6,000km and add the 400 km we get an orbital circumference of 40212km. This means that the satellite travels around the globe 500 times an hour or 12000 times a day! Where am I going wrong??

 

[kuruman]

That is excessive. Why is R = 400 km? What is R in the Law of Gravitation?

 

[max0005]

R is 400km because my textbook says (quote): "whose orbit is 400km above the Earth's surface". R stands for radius.

 

[kuruman]

R is 400km because my textbook says (quote): "whose orbit is 400km above the Earth's surface". R stands for radius.

What kind of radius? From where to where?

 

[max0005]

Well, I think it's the distance between the satellite and earth... But then book doesn't explain if I should consider from the center of the Earth or from the surface. Indeed, if it was from the center then I'd get a speed of about 2.5*10^5km/h.

 

[kuruman]

It is from the center of the Earth, i.e. the radius of the orbit, not the altitude of the satellite. Also don't forget to convert kilometers to meters before you calculate.

 

[max0005]

So my R would have a value of 6.4*10^5 meters right? Then I get a velocity of 25,000m/s or 25km/h which makes a lot more sense.

Thank you very much for your help!

 

[max0005]

A general question... Provided a know the radius of orbit I can always use:

$$v=\sqrt\frac{G(Mass of Earth)}{R}$$ and from there the period?

 

[kuruman]

No. The radius of the Earth is about 6400 km to which you need to add the 400 km of altitude. That is the radius of the orbit that needs to go in the calculation.

 

[kuruman]

A general question... Provided a know the radius of orbit I can always use:

$$v=\sqrt\frac{G(Mass of Earth)}{R}$$ and from there the period?

Yup.

 

[max0005]

Then if I had a satellite with mass 200kg, g at 8.2m/s^2 and total radius of 7*10^6 at would have a speed of 75km/h?

Edit: Sorry about the other, I plunged in wrong decimal. :\

 

[kuruman]

Then if I had a satellite with mass 200kg, g at 8.2m/s^2 and total radius of 7*10^6 at would have a speed of 75km/h?

Edit: Sorry about the other, I plunged in wrong decimal. :\

Is this a different problem? What is the statement of this one?

 

[max0005]

Considering a satellite orbiting with a radius of 7*10^6m from the center of Earth having a mass of 200kg. The value of g (gravity) is fixed at 8.2m/s. Estimate the speed of the satellite.

$$v=\sqrt\frac{G(MassEarth)}{7*10^6}$$

Which results in about 75.6km/h.

 

[kuruman]

Considering a satellite orbiting with a radius of 7*10^6m from the center of Earth having a mass of 200kg. The value of g (gravity) is fixed at 8.2m/s. Estimate the speed of the satellite.

$$v=\sqrt\frac{G(MassEarth)}{7*10^6}$$

Which results in about 75.6km/h.

I did not check the numbers, but you should use the same equation. I don't see, however, why you are given the mass and g at that orbit. Are there more parts to the problem?

 

[max0005]

Quote from book:

"Calculate the speed of a 200kg satellite, orbiting the Earth at a height of 7.0*10^6 meters.

Assume that g = 8.21 m/s^2 for this orbit."

 

[kuruman]

Quote from book:

"Calculate the speed of a 200kg satellite, orbiting the Earth at a height of 7.0*10^6 meters.

Assume that g = 8.21 m/s^2 for this orbit."

It is better to say a = g = v²/r and solve for v that way.

 

[max0005]

Is that the acceleration obtained from centripetal formula? I don't understand how it works then, one is the value of gravity due to the planet, the other is the speed of the satellite...

 

[kuruman]

$$F_{Net}=ma$$

$$F_{Net}=\frac{GmM_E}{R^2}=mg$$

$$F_{Net}=\frac{mv^2}{R}$$

What is g at a given orbit, what is a and what is the centripetal acceleration?