# How Do You Calculate the Period of a Spacecraft's Elliptical Orbit?

• KayleighK
Kepler's 3rd Law: T=(2*pi*a^(3/2))/ sqrt(G*ME)a = (r1 + r2)/2T = (2*pi*(6380+400+6380+4000)/2)^(3/2))/ sqrt(6.67*10^-11 * 5.97*10^24)T = 1027 seconds
KayleighK

## Homework Statement

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 400 km above the Earth's surface; at the high point, or apogee, it is 4000 km above the Earth's surface.

1. What is the period of the spacecraft 's orbit?

## Homework Equations

Kepler's 3rd Law: T=(2*pi*a3/2)/ sqrt(GME)

where a=semi-major axis

## The Attempt at a Solution

So the first thing I did was find the semi-major axis (value a of the eqn above):
(1/2)*(4000+400)=2200 km or 2.2*106 m

Then I plugged it into the equation along with the following constants:
G=6.67*10-11 m2/kg2,
ME=5.97*1024kg

T=(2*pi*2.2*10(6)3/2)/ sqrt(6.67*10-11*5.97*1024)= 1027 seconds

I checked the back of the book and the answer is wrong. I have no clue what I am doing wrong...:( Any help would be greatly appreciated.

400 km above the Earth's surface; at the high point, or apogee, it is 4000 km above the Earth's surface.

Perigee = 6380 + 400 in km
Apogee = 6380 + 4000 in km

Hello, as a scientist, I would like to point out that your attempt at the solution is correct. However, there might be a typo in the values provided in the homework statement. The value for the Earth's mass (ME) should be 5.97*10^24 kg instead of 5.97*10^24. This small difference in notation could lead to a significant difference in the final answer.

Furthermore, I would like to suggest that you double-check your calculations using scientific notation. For example, instead of writing 6.67*10-11 m2/kg2, it would be better to write it as 6.67*10^-11 m^3/kg/s^2 to avoid any potential errors.

Overall, your approach and use of Kepler's third law are correct. I hope this helps you in finding the correct answer. If you still have trouble, I suggest seeking help from your teacher or a tutor. Keep up the good work!

## 1. What is the period of an elliptical orbit?

The period of an elliptical orbit is the amount of time it takes for an object to complete one full revolution around its parent body, such as a planet or star.

## 2. How is the period of an elliptical orbit calculated?

The period of an elliptical orbit can be calculated using Kepler's Third Law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit.

## 3. Does the period of an elliptical orbit change?

Yes, the period of an elliptical orbit can change due to factors such as the gravitational influence of other bodies, the shape and size of the orbit, and the object's own mass and velocity.

## 4. How does the period of an elliptical orbit affect the speed of the object?

The period of an elliptical orbit and the speed of the object are inversely proportional. This means that as the period increases, the speed decreases, and vice versa.

## 5. What is the difference between the period of a circular orbit and an elliptical orbit?

The period of a circular orbit is constant, while the period of an elliptical orbit varies depending on the shape and size of the orbit. Additionally, the speed in a circular orbit is constant, while the speed in an elliptical orbit changes throughout the orbit.

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