# Calculating Gravity and Weight on Krypton

• Cicima
In summary, The weight on Krypton for a person with a mass of 70 kg can be calculated using the equation Fg = Gm1m2 / r^2, where m1 is the mass of the planet, m2 is the person's mass, G is the gravitational constant, and r is the distance from the center of the planet to its moon. After setting up the equation, it is important to plug in the correct values for each variable and solve algebraically to find the mass of the planet, which is then multiplied by the person's mass and divided by the radius squared to calculate weight. It is important to note that r squared is not the same as r multiplied by 2.
Cicima
Im not sure if this is done right...

The planet Krypton has a moon that circles the planet in a circle that has a period of 40 days (watch units) and is 6 x 10^8 m from the planet's center. The planet's radius is 6.4 x 10^6 m. What would you weigh on Krypton if your mass is 70 KG? Show a free body diagram.

Im guessing use Net force = ma
The net force is the Gravitational force, Gm1m2 / r^2 is = m2 4 (pie)^2 r/ T^2

for the 4 (pie)^2 r/ T^2 r is the distance form the planet's center, to the moon, for T, find how many second 40 days is.

and the other r is the planet's radius

m2 cancels

solve for m1, and yo uhave the mass of the planet.

I guess plug in numbers after that, into the equaltion Fg = Gm1 m2 / r^2

m2 would be the 70 kg

Can someone help me out and tell me if what i am doing looks right?

Cicima said:
Im not sure if this is done right...

Im guessing use Net force = ma
The net force is the Gravitational force, Gm1m2 / r^2 is = m2 4 (pie)^2 r/ T^2

for the 4 (pie)^2 r/ T^2 r is the distance form the planet's center, to the moon, for T, find how many second 40 days is.

and the other r is the planet's radius ?

m2 cancels

solve for m1, and yo uhave the mass of the planet.

I guess plug in numbers after that, into the equaltion Fg = Gm1 m2 / r^2

m2 would be the 70 kg

Can someone help me out and tell me if what i am doing looks right?
All the r in your first equation are the same. r is the distance from the center of Krypton to the center of its moon. The r in the second equation is the radius of the planet. I assume that Fg is the force of gravity, or weight at the planet's surface. If that is what you mean, then you should have the correct result.

Yes that's what i meant. The other r (the one in the 2nd equation) is the radius. So i know how to set it up, the problem is I am not good at plugging it in. What am i doing wrong?

Gm1m2 / r^2 is = m2 4 (pie)^2 r/ T^2
for the 4 (pie)^2 r/ T^2 r is the distance form the planet's center, to the moon, for T, find how many second 40 days is.
m2 cancels
so it would be...

m1/3.6x10^17=39.47 r/t^2
m1/3.6x10^17=39.47 (6x10^8)/1.194x10^13
when i solve for m1 i get 6.84x10^14...that doesn't seem right?

Fg = Gm1 m2 / r^2

m2 would be the 70 kg
and if its right, how do u multiply that times 70 kg...which is m2?

So am i wrong? And if so, where am i plugging them in wrong?

Cicima said:
Yes that's what i meant. The other r (the one in the 2nd equation) is the radius. So i know how to set it up, the problem is I am not good at plugging it in. What am i doing wrong?

m1/3.6x10^17=39.47 r/t^2
m1/3.6x10^17=39.47 (6x10^8)/1.194x10^13
when i solve for m1 i get 6.84x10^14...that doesn't seem right?

and if its right, how do u multiply that times 70 kg...which is m2?
It doesn't look right, but you have not included enough steps to find out what you are doing. I suggest you solve the equation algebraically for m1 before entering any numbers,

OlderDan said:
It doesn't look right, but you have not included enough steps to find out what you are doing. I suggest you solve the equation algebraically for m1 before entering any numbers,

m1/r^2=4pi^2 r/t^2
thats basically as far as i can solve it...
plug in 345600 for t, and 6x10^8 for r, then its just basic calculator plug ins from there getting m1/r^2=.0019...so what am i doing wrong?

Cicima said:
m1/r^2=4pi^2 r/t^2
thats basically as far as i can solve it...
plug in 345600 for t, and 6x10^8 for r, then its just basic calculator plug ins from there getting m1/r^2=.0019...so what am i doing wrong?
What became of G? Why not multiply both sides by r² ?

OlderDan said:
What became of G? Why not multiply both sides by r² ?
Oops forgot G.

Gm1=4pi^2 r^3/t^2

That would be simplified then...so why does my answer look so wrong when i plug in 345600 for t, and 6x10^8?

Cicima said:
Oops forgot G.

Gm1=4pi^2 r^3/t^2

That would be simplified then...so why does my answer look so wrong when i plug in 345600 for t, and 6x10^8?

Then...
Gm1=39.478 r^3/3456000^2
Gm1=39.478 r^3/1.194x10^13
then you divide each side by G...so what am i doing wrong? Maybe I am taking r^3 wrong? How do you do that just to make sure I am doing it right.

Cicima said:
Oops forgot G.

Gm1=4pi^2 r^3/t^2

That would be simplified then...so why does my answer look so wrong when i plug in 3456000 for t, and 6x10^8?

Gm1=4pi²*r³/t²
m1=4*pi²*r³/(t²*G) = 1.07 x 10^25 kg on my calculator

Edit: The original value posted was incorrect because it used the OP's value for t that was off by a factor of 10. The value in the quote has been corrected, and the correction for m1 made here

Last edited:
1.07 x 10^27 kg now needs to be multiplied by m12 which is 70kg. so can i just multiply 1.07x10^27 by 70, then divide it by r^2 and that's it?

Quick question, if I am doing R^2, would that mean if r is 6.4x10^6 that r^2 is 6.4x10^12?

OlderDan said:
Gm1=4pi²*r³/t²
m1=4*pi²*r³/(t²*G) = 1.07 x 10^27 kg on my calculator
can u show me the steps because i still can't get that answer, after that i know how to do it...thanks.

Cicima said:
Quick question, if I am doing R^2, would that mean if r is 6.4x10^6 that r^2 is 6.4x10^12?
NO

(6.4x10^6)^2 = 6.4x10^6x6.4x10^6 = (6.4)^2x10^12 = 40.96x10^12 = 4.096x10^13

can u show me the steps because i still can't get that answer...i get like 7.139x10^13, then i multiply it by 70 and 9.8 and divide it by (6.4x10^6)^2 and get 1195. Howd u get 1.07x10^27? What do you do after you get it? if u plug it into the 2nd equation you get a crazy number.

Cicima said:
can u show me the steps because i still can't get that answer...i get like 7.139x10^13, then i multiply it by 70 and 9.8 and divide it by (6.4x10^6)^2 and get 1195. Howd u get 1.07x10^27? What do you do after you get it? if u plug it into the 2nd equation you get a crazy number.
t = 40days*(24hr/day)*(60min/hr)*(60sec/min) = 3.456x10^6 s <== You had this as 3.456x10^5 and I used your number, so my exponent was a bit off

t² = (3.456x10^6 s)(3.456x10^6 s) = 3.456²x10^12 s² = 1.194x10^13 s²

t²*G = (1.194x10^13 s²)*(6.673x10^-11 N m²/kg²) = 7.97x10² N m² s²/kg² = 7.97x10² m³/kg

m1 = 4*pi²*r³/(t²*G) = 4*3.14²*(6x10^8 m)³/(7.97x10² m³/kg)

m1 = 4*9.870*6³x10^24 m³/(7.97x10² m³/kg)

m1 = 8.527 x10^27 kg/(7.97x10²)

m1 = 1.07 x10^25 kg

Last edited:

## 1. How is gravity calculated on Krypton?

Gravity on Krypton is calculated using the same formula as on Earth: F = Gm1m2/d^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and d is the distance between them.

## 2. How does gravity on Krypton compare to Earth?

Gravity on Krypton is approximately 3.7 times stronger than on Earth. This means that objects will weigh almost four times as much on Krypton compared to their weight on Earth.

## 3. What is the weight of an average person on Krypton?

The weight of an average person on Krypton would vary depending on their mass, but on average, a person would weigh about four times their weight on Earth.

## 4. How does the gravity on Krypton affect objects?

The stronger gravity on Krypton means that objects will fall faster and have more weight compared to on Earth. This could also affect the structural integrity of buildings and other structures on Krypton, as they would need to be built to withstand the increased gravitational force.

## 5. Can gravity on Krypton be altered or manipulated?

In theory, gravity on any planet or celestial body can be altered or manipulated through advanced technology or natural phenomena. However, there is currently no known way to alter gravity on Krypton.

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