# Homework Help: Gravitational field strength at a point

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1. Jul 18, 2016

### Zack K

1. The problem statement, all variables and given/known data
What is the gravitational field strength at a point 6.38x106 m above the earth's surface?

2. Relevant equations

3. The attempt at a solution
Ok so I plug in all the known variables. G= 6.67x10-11. M= 5.98x1024. r= 6.38x106 and so I multiply that by 2 because the given point is the exact same number as r. Then I powered by two. I calculate both top an bottom of the equation to get g= 39.8866x1013/162.8176x1012. For the answer I get a field strength of 2.449 m/s2. But the answer says that it is 0.622 m/s2.

Edit. I was looking at a wrong answer. You got to love my brain sometimes!

2. Jul 18, 2016

### Andrew Mason

There is an easy way to do this if you recognize what 6,380 km above the surface is in terms of the earth radii!

AM

3. Jul 19, 2016

### Zack K

Isn't 6,380 km above the surface just twice the radius of Earth?

4. Jul 19, 2016

### Andrew Mason

Right. So how should acceleration due to gravity there compare to the acceleration on the surface?

AM

5. Jul 19, 2016

### Zack K

Oh I see. So it would approximately be the square root of 9.8?

6. Jul 19, 2016

### Andrew Mason

? Gravitational force and Gravitational force/unit mass (acceleration) is proportional 1/R2 so if R doubles from the surface, what happens to acceleration (ie. compared to acceleration at the surface, which is 9.8 m/sec2)?

AM

7. Jul 20, 2016

### haruspex

This illustrates a benefit of always including units. In this case, the square root of 9.8m/s2 would be about $3.1m^{\frac 12}s^{-1}$. Hmm.. I wonder what the square root of a metre looks like.

8. Jul 20, 2016

### Zack K

It would diminish by 4x?

9. Jul 20, 2016