NASA decides to put a 128 kg satellite into orbit over the planet Venus to take pictures. The satellite is 37,000,000 m above the surface of Venus.

Calculate the gravitational field at that altitude in m/s². How do you calculate this?

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 cente**r 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 = 0.2373 m/sec²

Answered by AL2006 | 2024-06-10

The gravitational field strength at an altitude of 37,000,000 meters above Venus is approximately 0.175 m/s². This is calculated using the surface gravity of Venus and applying the inverse square law based on the distance from the center of the planet. Consequently, gravity decreases as you move further away from the planet's center.
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Answered by AL2006 | 2025-02-13