# Centripetal acceleration of the satellite?

• magnolia1
In summary, a satellite in a circular orbit 790 km above the Earth's surface has a period of 100.5 min. To find its speed, the radius of the orbit must be calculated, which is 7.16*10^6 meters. Substituting this value and the given period into the formula v=(2 pi r)/T, the speed is calculated to be 7461 m/s.
magnolia1

## Homework Statement

An Earth satellite moves in a circular orbit 790 km above the Earth's surface. The period of the motion is 100.5 min.
(a) What is the speed of the satellite?
(b) What is the magnitude of the centripetal acceleration of the satellite?

a= v2 / r

T = (2 Pi r)/ V

## The Attempt at a Solution

Converting km to m: 790km*(103m / 1km) = 7.90*105m
Converting min to seconds: 100.5min*(60seconds/1min) = 6.03*103seconds

Now that I have the conversions, I will solve for the unknown variables in the above formulas:
I think the radius is 7.90*105m and period is 6.03*103seconds

(a) Speed of satellite:
v=(2 pi (7.90*105m)) / (6.03*103s)
v=8.27*102m/s

because the speed is incorrect, the error follows through to part (b) when substituting in speed.
How is it that it's so far off?
Thanks :)

r should be the radius of the orbit, centred at the centre of Earth. The height of the satellite above the surface of Earth is 790 km, how far is it from the centre of Earth?

ehild

1 person
I had to look it up, but 6371 km is from the surface of Earth to its centre. So I will try now with the new radius of (6371km+790km) * (10^3m/1km) = 7.16*10^6metres.
ANS: 7460.63 = 7461m/s

magnolia1 said:
I had to look it up, but 6371 km is from the surface of Earth to its centre. So I will try now with the new radius of (6371km+790km) * (10^3m/1km) = 7.16*10^6metres.
ANS: 7460.63 = 7461m/s

You are welcome.

ehild

As a scientist, it is important to carefully check and review your calculations to ensure accuracy. In this case, the error may be due to rounding off numbers too early in the calculation or using the incorrect values for the radius and period.

Let's review the equations used:

a = v^2 / r

T = (2πr) / v

In the first equation, the radius (r) should be the distance from the center of the Earth to the satellite's orbit, which is the sum of the Earth's radius (6.37*10^6 m) and the satellite's altitude (790 km or 7.90*10^5 m). So the radius (r) should be 7.97*10^6 m.

In the second equation, the period (T) should be in seconds. So the period (T) should be 100.5 min * (60 seconds / 1 min) = 6030 seconds.

Now let's plug in the correct values:

(a) Speed of satellite:
v = (2π * 7.97*10^6 m) / (6030 s)
v = 8.29*10^3 m/s

(b) Magnitude of centripetal acceleration:
a = (8.29*10^3 m/s)^2 / (7.97*10^6 m)
a = 8.59*10^-2 m/s^2

(a) Speed of satellite: 8290 m/s
(b) Magnitude of centripetal acceleration: 0.0859 m/s^2

It is important to double check your calculations and use accurate values for the variables in your equations to ensure the correct answer. Additionally, using scientific notation can help avoid errors and provide more accurate results.

## What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It is equal to the square of the object's speed divided by the radius of the circle.

## How is centripetal acceleration related to the satellite's orbit?

Centripetal acceleration is an essential component of the satellite's orbit. It is responsible for keeping the satellite in its circular path around the Earth, balancing out the gravitational pull of the Earth.

## What factors affect the magnitude of centripetal acceleration for a satellite?

The magnitude of centripetal acceleration for a satellite depends on the satellite's speed, the radius of its orbit, and the mass of the object it is orbiting around (e.g. the Earth).

## How is centripetal acceleration calculated for a satellite in orbit?

Centripetal acceleration can be calculated using the formula a = v²/r, where a is the centripetal acceleration, v is the velocity of the satellite, and r is the radius of its orbit. This can also be expressed as a = ω²r, where ω is the angular velocity of the satellite.

## How does centripetal acceleration affect the stability of a satellite's orbit?

Centripetal acceleration is crucial in maintaining the stability of a satellite's orbit. If the centripetal acceleration is too low, the satellite will drift away from its orbit, and if it is too high, the satellite will spiral towards the Earth. Therefore, the magnitude of centripetal acceleration must be carefully controlled for a stable orbit.

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