- #1

Gebri Mishtaku

- 19

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So I found this problem in my 10th grade Physics workbook and since its test time tomorrow I must be able to work everything out. The problem I'm having here is just the calculations which give me a time short of one hour from the answer on the book. Anysways, please have a go:

An astronaut is flying on a shuttle that's on the same orbit(400 km above the surface of Earth) as a satellite that he needs to repair. He is 50 km behind the satellite and in order to reach it, the astronaut lowers his orbiting radius by 2 km. What is the time needed for the astronaut to reach the satellite?

It doesn't say anything about it being an aerial distance, so it must be taken to be circular.

G=6.67x10^-11Nm^2/kg^2

M

R

The equation for the Law of Universal Attraction must be used and also that of the centripetal force and acceleration in circular motion:

F

I tried going around this problem calculating the relative linear speed of the astronaut to the satellite and then the time of approach would just be the ratio of the distance the astronaut had to travel relative to the satellite(50km) with the astronaut's relative linear velocity. I know I'm missing something here because my calculations(which in this problem are loathsome to be honest) come out ≈ 1.5 hours less than the answer on the book, which wouldn't surprise me if it was wrong because this book has plenty of mistakes in it.

Formula and answer are found in the attachment.

## Homework Statement

An astronaut is flying on a shuttle that's on the same orbit(400 km above the surface of Earth) as a satellite that he needs to repair. He is 50 km behind the satellite and in order to reach it, the astronaut lowers his orbiting radius by 2 km. What is the time needed for the astronaut to reach the satellite?

It doesn't say anything about it being an aerial distance, so it must be taken to be circular.

G=6.67x10^-11Nm^2/kg^2

M

_{Earth}=6x10^24kgR

_{Earth}=6.4x10^6m## Homework Equations

The equation for the Law of Universal Attraction must be used and also that of the centripetal force and acceleration in circular motion:

F

_{c}=mv^2/r and F_{G}=(Gm_{1}m_{2})/r^2## The Attempt at a Solution

I tried going around this problem calculating the relative linear speed of the astronaut to the satellite and then the time of approach would just be the ratio of the distance the astronaut had to travel relative to the satellite(50km) with the astronaut's relative linear velocity. I know I'm missing something here because my calculations(which in this problem are loathsome to be honest) come out ≈ 1.5 hours less than the answer on the book, which wouldn't surprise me if it was wrong because this book has plenty of mistakes in it.

Formula and answer are found in the attachment.

#### Attachments

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