# Diameter of Earth-like planet / gravitational force

• chris2112
In summary, the radius of the Earth would need to be about 1.3 times bigger in order to make the average person weigh 50 more pounds.
chris2112
I kind of have a stupid question. How much more massive would the Earth have to be for the average person to weigh about 50 more pounds? How much bigger could the Earth's diameter be? Thanks.

There isn't a single number. Decreasing the radius of the Earth while keeping the mass the same would increase the weight of a person on the surface. You could also keep the Earth the same radius but increase the mass to get the same effect. Or any combination of the two.

chris2112 said:
I kind of have a stupid question. How much more massive would the Earth have to be for the average person to weigh about 50 more pounds? How much bigger could the Earth's diameter be? Thanks.

Let's say average weight for women is 145 pounds and for men 185 pounds and average "person" weighs 165 pounds.

Lets say "earth-like" means roughly same DENSITY because roughly same chemical composition---iron core, rocky mantle, crust etc etc.

So you want the size for which the 165 pound person will weigh 50 more pounds, or 215. So you want gravity to be stronger by a factor of 215/165.

So you want the RADIUS to increase by that same factor.

So calculate what 215/165 is. About 1.3.

So with those assumptions (about "average person" and "earth-like") the answer is that the radius would need to be about 30% bigger.
IOW the diameter would need to be about 30% bigger.

Last edited:
marcus, I am not following your logic. Ifd the radius increased by 30%, the mass would increase by 2.2. Are you accounting for that?

DaveC426913 said:
marcus, I am not following your logic. Ifd the radius increased by 30%, the mass would increase by 2.2. Are you accounting for that?
You'd also be 30% farther away from the centre of the gravitational field. Marcus is correct.

Last edited:
TEFLing
G ~ M/R^2
Surface gravityp ~ M/R^3
Density

G ~ p x R

marcus
True, TEFL. The surface gravity is proportional to the density times the radius.

BTW at the top of the blank space where we write replies there is this line of symbols B I U ... Σ
and if you click on the the Σ you get a convenient menu of symbols including the lowercase Greek rho (ρ) which is normally used for density.

There's also an x2 symbol you can click on to write superscripts. So your second equation could be written ρ ~ M/R3

You may have discovered this already, but in case not I wanted to mention it. Very handy.

I do write the ^ for superscripts when I want to use the google calculator, to be able to paste the formula into the google window and have it evaluate it. The calculator likes to have exponents written in the x^2 form. But for purely human consumption it looks better to use the PF symbols gadget.

Last edited:
TEFLing

## 1. What is the diameter of an Earth-like planet?

The diameter of an Earth-like planet can vary, but it is generally around 12,742 kilometers (7,917 miles). This is about 1.7 times smaller than the diameter of Earth.

## 2. How does the diameter of a planet affect its gravitational force?

The diameter of a planet does not directly affect its gravitational force. However, a larger diameter can lead to a higher mass, which in turn can increase the gravitational force between the planet and other objects.

## 3. What is the gravitational force of an Earth-like planet?

The gravitational force of an Earth-like planet depends on its mass and the mass of other objects around it. On average, the gravitational force on the surface of an Earth-like planet is 9.8 meters per second squared.

## 4. Can the gravitational force on an Earth-like planet change?

Yes, the gravitational force on an Earth-like planet can change if its mass or the mass of other objects around it changes. For example, if a large asteroid were to collide with the planet, the gravitational force may increase.

## 5. How is the gravitational force on an Earth-like planet calculated?

The gravitational force on an Earth-like planet is calculated using Newton's Law of Universal Gravitation, which states that the force is proportional to the masses of the two objects and inversely proportional to the square of the distance between them. This can be represented by the equation F = (G * m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

### Similar threads

• Astronomy and Astrophysics
Replies
17
Views
2K
• Astronomy and Astrophysics
Replies
5
Views
776
• Astronomy and Astrophysics
Replies
1
Views
1K
• Astronomy and Astrophysics
Replies
39
Views
3K
• Astronomy and Astrophysics
Replies
21
Views
1K
• Astronomy and Astrophysics
Replies
15
Views
1K
• Astronomy and Astrophysics
Replies
2
Views
2K
• Astronomy and Astrophysics
Replies
19
Views
1K
• Astronomy and Astrophysics
Replies
2
Views
1K
• Astronomy and Astrophysics
Replies
12
Views
1K