How do i find gravity given a radius and an altitude?

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To calculate the acceleration due to gravity at an altitude of 1.00 x 10^6 meters above Earth's surface, the formula g = GM/(R+h)^2 is used, where R is the Earth's radius and h is the altitude. The gravitational constant G is approximately 6.673 x 10^-11, and the mass of the Earth can be derived from the known acceleration due to gravity at the surface, which is about 9.8 m/s^2. It's crucial to ensure all units are consistent, avoiding mixing different measurement systems. The final calculation yields a gravity value of approximately 7.33 m/s^2 at the specified altitude. Understanding these principles is essential for solving similar physics problems effectively.
gcombina
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Homework Statement



What is the acceleration due to gravity at an altitude of 1.00 x 10^6 above the Earth's surface, given that the radius of the Earth is 6.38 x 10^6 m?
How do i go about solving that?

Homework Equations


Using g = Gm/r^2

The Attempt at a Solution


g= GMm/(R+h)^2
R= radio of earth
H= height/altitude given

so

g = (6.67300 x 10^-11) (m) / [(6.38 x 10^6 m) + (1.00 x 10^6)]^2

*** my question is, what do I put as M and m? **
 
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gcombina said:

Homework Statement



What is the acceleration due to gravity at an altitude of 1.00 x 10^6 above the Earth's surface, given that the radius of the Earth is 6.38 x 10^6 m?
How do i go about solving that?

Homework Equations


Using g = Gm/r^2

The Attempt at a Solution


g= GMm/(R+h)^2
R= radio of earth
H= height/altitude given

so

g = (6.67300 x 10^-11) (m) / [(6.38 x 10^6 m) + (1.00 x 10^6)]^2

*** my question is, what do I put as M and m? **
In your Relevant equations your formula for gravitational acceleration contains only one "m", and it represents the mass of the body that is causing the acceleration. In this case what is the object?

In your problem statement the "m" on the radius of the Earth is the units: m for meters. There should be units associated with the altitude figure, too. What are they?

It is very important to keep the units associated with values. You don't want to be mixing miles with centimeters in a calculation! Instructors will deduct marks if units are left off of results.
 
gcombina said:

Homework Statement



What is the acceleration due to gravity at an altitude of 1.00 x 10^6 above the Earth's surface, given that the radius of the Earth is 6.38 x 10^6 m?
How do i go about solving that?

Homework Equations


Using g = Gm/r^2

The Attempt at a Solution


g= GMm/(R+h)^2
R= radio of earth
H= height/altitude given

so

g = (6.67300 x 10^-11) (m) / [(6.38 x 10^6 m) + (1.00 x 10^6)]^2

*** my question is, what do I put as M and m? **
You need not put M, actually the acceleration due to gravity on Earth is well known constant it is g=9.8 m/s^2
so put this value for GM/R^2 so you can calculate value of GM as you know radius of Earth R for acceleration due to gravity at height h now you calculate the Force on unit mass i.e. m =1 kg then you get result for value of g at an altitude.(actually Force per unit mass is acceleration; in your expression given above you have erroneously written g istead of force using Newton's law of gravitation
 
Gravity at altitude of 10^6 m
Gh = Go(re/(re+h))^2
Go = 9.80665 m/s^2
re = 6367444.7 m
h = 1,000,000 m
Gh = 7.33 m/s^2
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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