What is the weight of a 22.0 kg mass at the equator on the surface of Saturn? The mass of Saturn is 5.69 x 10^26 kg and the radius of Saturn is 6.03 x 10^7 m

I tried this one using the equation :
Fg= Gm1m2/r^2
Fg= (6.67 x 10^-11)(22.0kg)(5.69 x 10^26kg) / (6.03 x 10^7m)^2
I got the answer 230 N Is this correct?

Yes it is correct

No its not its 219

If anyone comes across this, don't listen to the second Anonymous answer. 230N is the correct answer, not 219.

I got 225.5930309

Nevermind i made a mistake. 230N is correct

To determine the weight of a 22.0 kg mass at the equator on the surface of Saturn, we can use the equation for gravitational force:

Fg = (G * m1 * m2) / r^2

where:
Fg is the gravitational force,
G is the gravitational constant (approximately 6.67 x 10^-11 N*m^2/kg^2),
m1 is the mass of the first object (in this case, the mass of the 22.0 kg mass),
m2 is the mass of the second object (in this case, the mass of Saturn), and
r is the distance between the centers of the two objects (in this case, the radius of Saturn).

Let's plug in the values:

Fg = (6.67 x 10^-11 N*m^2/kg^2) * (22.0 kg) * (5.69 x 10^26 kg) / (6.03 x 10^7 m)^2

Calculating this expression will give us the answer:

Fg = 230 N

So, your calculation is correct. The weight of a 22.0 kg mass at the equator on the surface of Saturn is approximately 230 N.