1). Look up the acceleration of gravity on the surface of Venus. It's 8.87 m/s². (That's about 10% less than on Earth.)
2). Look up the radius of Venus. It's 6051.8 km. (That's about 5% less than Earth's.)
3). Remember that the gravitational field (acceleration, force) changes opposite to the square of the distance from the planet ... or the distance between any two masses that are gravitating.
Now we have enough information to do the calculating. Notice that the question only asks for the planet's "gravitational field" (acceleration) way out there. That has nothing to do with the satellite's mass, or whatever you decide to put out there, or even if there's nothing out there at all. It's just a characteristic of Venus at that distance from it.
The distances we need to compare are the distances from the center of Venus.
On the surface (distance from the center is the radius of Venus), it's 6,051,800 m. In the orbit, it's 37,000,000 m.
The ratio is (37,000,000 / 6,051,800) = about 6.1 .
The gravity way out in the orbit is less than on the surface, and by the same amount as the square of the distance ratio.
Surface gravity = 8.87
Gravity out at the orbit = 8.87 divided by (the distance ratio)²
Gravity = 8.87 / (6.1)²
Better way: Gravity = 8.87 x (6,051,800/37,000,000)² = <em>0.2373 m/sec²</em>
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
