Planet Kling has half the radius and 2 times the mass of the Earth. What is the best estimate for the magnitude of the gravitational field at the surface of planet Kling?

80m/s²

Explanation:

First, use the equation below to find the strength of a planet/moon's satellite field and acceleration rate.

This is G × Mass of planet/moon ÷ r² (which is the distance or conversely radius squared)

Then you want to plg in all your values, which are going to be the gravitational constant, Kling's mass, and Kling's radius.

Remember that Earth's mass is equal to 5.97 × 10∧24 kg and its radius is exactly 6,731 km, so this will help you figure out your values for any other problems like this .

Also that the big ( G) you see is = 6.673×10∧-11.

Now for step two, which is the last thing you need to do. Once you've simplified the above equation, take the value you found, which should ≈ 7.85, and then multiply it by the force of gravity on Earth.

Once you multiply 7.85 by 9.8 m/s², the answer will be = 76.93 or ≈ 80m/s².

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )