Gravity on Earth: 70% Value Above Surface?

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To determine how far above Earth's surface gravity is reduced to 70% of its surface value, the gravitational force equation can be utilized, focusing on the relationship between distance and gravitational force. The gravitational force decreases with the square of the distance from the Earth's center, meaning that as one moves away, the force diminishes according to the inverse square law. Specifically, if the distance from the Earth's center is doubled, the gravitational force becomes one-fourth of its original value. To find the specific height above the surface that results in 70% of the gravitational force, the Earth's radius must be considered, as it plays a crucial role in calculating the required distance. Ultimately, understanding this relationship allows for the determination of how far one must ascend to achieve the desired reduction in gravitational force.
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Homework Statement



How far above the surface of the Earth do you need to be for the force of gravity to be reduced to 70% of its value on the surface.

Homework Equations


g on surface is 9.8 m/s^2

Fg=G*m1*m2/r^2


The Attempt at a Solution



Is there a way to do this without knowing the mass of the earth? It seems as though the gravity equation I have above could only be useful if you knew either the mass of the Earth or the radius of the earth. I was just wondering if there is a way to approach this without knowing either of those. Thanks!
 
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Instead of plug and chug into the gravitational force equation, try working with ratios.
Everything drops out except for the radii between the masses.
 
Since you are being asked for the specific distance, then you will ultimately need to know the Earth's radius, which is easy enough to find.

Remember that "r" is not zero at Earth's surface, but "R" (let that represent the Earth's radius.

Think about the "inverse square proportionality" of force with distance. If distance doubled (that is, if you went a distance of R above the surface so the distance from Earth's center is "2R"), the field and therefor the force would be (1/4)F , or (0.25)F (compared to F on the surface). If distance was 3R, the force would be (1/9)F , or (0.111)F .

You need to figure out what factor of R would produce (0.70)F
 

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