Find the force of attraction on a 3 kg mass at the surface of a planet given the following information about the planet:

mass = 5.98 x 10^24 kg, radius = 6.38 x 10^6 m.

Find the force if the radius is doubled to 12.76 x 10^6m

Thanks!!

F = G m M /r^2

where
G = 6.67 * 10^-11

If you double the radius, you divide the force by 4 (1/2*1/2) = 1/4

29.4 N

for part b

7.35 N

To find the force of attraction on a mass at the surface of a planet, we can use the equation for gravitational force:

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

where F is the force of attraction, G is the gravitational constant (6.673 x 10^-11 N m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.

In this case, one of the objects is the planet and the other object is the 3 kg mass. The mass of the planet is given as 5.98 x 10^24 kg, and the radius of the planet is given as 6.38 x 10^6 m.

Plugging these values into the equation, we get:

F = (6.673 x 10^-11 N m^2/kg^2) * ((3 kg) * (5.98 x 10^24 kg)) / (6.38 x 10^6 m)^2

Simplifying the equation, we get:

F = (6.673 x 10^-11 N m^2/kg^2) * (17.94 x 10^24 kg) / (40.64 x 10^12 m^2)

F = (6.673 x 10^-11 N m^2/kg^2) * (17.94 / 40.64) x (10^24 kg / 10^12 m^2)

F = (6.673 x 10^-11 N) * (0.4419) x (10^12)

F = 2.946 x 10^-11 N

So, for the given information, the force of attraction on the 3 kg mass at the surface of the planet is approximately 2.946 x 10^-11 N.

Now, if the radius of the planet is doubled to 12.76 x 10^6 m, we can use the same equation to find the new force of attraction.

Plugging in the new radius value, we get:

F_new = (6.673 x 10^-11 N m^2/kg^2) * ((3 kg) * (5.98 x 10^24 kg)) / (12.76 x 10^6 m)^2

Following the same steps as before, we find:

F_new = 7.365 x 10^-12 N

So, if the radius of the planet is doubled, the new force of attraction on the 3 kg mass at the surface of the planet would be approximately 7.365 x 10^-12 N.