Gravitational field strength at a point

[Zack K]

Homework Statement

What is the gravitational field strength at a point 6.38x10⁶ m above the Earth's surface?

Homework Equations

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!

 

[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

 

[Zack K]

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

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

 

[Andrew Mason]

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

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

AM

 

[Zack K]

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

AM

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

 

[Andrew Mason]

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

? 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

 

[haruspex]

the square root of 9.8?

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.

 

[Zack K]

? 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

? 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

It would diminish by 4x?

 

[Andrew Mason]

It would diminish by 4x?

Correct. Work that out and compare that to your answer.

AM

 

[Zack K]

Correct. Work that out and compare that to your answer.

AM

I actually figured this out when I had my test today it was one of the questions. I just put in some random masses and radius in the equation and saw that if you half the radius then the force of gravity increases by 4x.