# How did the textbook get this answer? (satellite motion question)

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The conversation is about a group of friends discussing their plans for the weekend. One person suggests going to the beach, another suggests going on a hike, and the third person suggests having a game night at home.In summary, the group of friends is deciding on plans for the weekend. One person suggests going to the beach, another suggests a hike, and the third person suggests a game night at home.
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Homework Statement
Suppose we want to place a weather satellite into a circular orbit 300 km above the earth's surface. The earth's radius is 6380 km= 6.38E6 m, and its mass is 5.98E24 kg
A. What speed, period, and radial acceleration must it have?
B. Find the speed and orbit radius for an earth satellite that has a period of 1 day (86400 s).
Relevant Equations
(all answers are rounded) I understand how to get part A as
v= sqrt((6.67E-11*5.98E24)/(3.0E5+6.38E6))= 7730 m/s
T= 2π(6.68E6)/7730= 5430 s
a= (7730^2)/6.68E6= 8.95 m/s^2
I don't know what I'm doing wrong, but I'm not getting the same answer as the textbook for part B (v= 3.07E3 m/s; r= 4.23E7 m). Can someone please explain how to do part B?
Below is my attempt at the problem:
T= 86400= 2πr/v
v= 2π(6.68E6)/86400= 4.86E7 m/s
r= (86400)(7730)/2π= 1.06E8 m

It seems to me that you are using the Earth's radius as the satellite's orbit radius for part B. You are correct in your approach to part A. I suggest doing the same in part B.

take your result from part A (namely ##v = \sqrt{\frac{GM}{r}}##) and plug it into ##T = \frac{2 \pi r}{v}## to get

##T = 2 \pi r \sqrt{\frac{r}{GM}}## solve for ##r## and then solving for ##v## is straight forward.

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All hail PhDeezNutz!

SammyS, berkeman and PhDeezNutz

## 1. How did the textbook calculate the orbital speed of the satellite?

The textbook likely used the formula for orbital speed, which is derived from setting the gravitational force equal to the centripetal force. The formula is $$v = \sqrt{\frac{GM}{r}}$$, where $$G$$ is the gravitational constant, $$M$$ is the mass of the Earth, and $$r$$ is the radius of the orbit (distance from the center of the Earth to the satellite).

## 2. How did the textbook determine the altitude of the satellite's orbit?

The altitude of the satellite's orbit can be found by subtracting the Earth's radius from the orbital radius. If the textbook provides the orbital radius $$r$$, the altitude $$h$$ is given by $$h = r - R$$, where $$R$$ is the Earth's radius.

## 3. How did the textbook find the period of the satellite's orbit?

The period of the satellite's orbit can be calculated using Kepler's third law, which relates the orbital period $$T$$ to the orbital radius $$r$$. The formula is $$T = 2\pi \sqrt{\frac{r^3}{GM}}$$, where $$G$$ is the gravitational constant and $$M$$ is the mass of the Earth.

## 4. How did the textbook derive the gravitational force acting on the satellite?

The gravitational force acting on the satellite can be found using Newton's law of gravitation: $$F = \frac{GMm}{r^2}$$, where $$G$$ is the gravitational constant, $$M$$ is the mass of the Earth, $$m$$ is the mass of the satellite, and $$r$$ is the distance between the centers of the Earth and the satellite.

## 5. How did the textbook explain the relationship between orbital speed and altitude?

The textbook likely explained that the orbital speed decreases with increasing altitude. This is because the gravitational force weakens with distance from the Earth, requiring a lower speed to maintain a stable orbit. The relationship is given by $$v = \sqrt{\frac{GM}{r}}$$, indicating that as $$r$$ (orbital radius) increases, the orbital speed $$v$$ decreases.

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