what is the gravitational field strength of mars measured at a position 400 km above its surface? mars has a mass of 6.39x10^23 kg and a radius of 3390 km.​

Correct me if I'm wrong.
G=GM/r^2
G=(6.67×10^-11)(6.39×10^23)/(3390000+400000)^2
G=(6.67×10^-11)(6.39×10^23)/(3790000)^2
G=2.967 m/s^2

looks reasonable but did not do the arithmetic.

To find the gravitational field strength of Mars at a position 400 km above its surface, you can use the formula for gravitational field strength:

g = G * (M / r^2)

Where:
g = gravitational field strength
G = gravitational constant (6.67 × 10^-11 N m^2/kg^2)
M = mass of Mars (6.39 × 10^23 kg)
r = distance from the center of Mars (radius of Mars + 400 km)

Let's calculate it step by step:

1. Calculate the total radius by adding the radius of Mars (3390 km) and the distance from its surface (400 km):
Total radius = 3390 km + 400 km = 3790 km

2. Convert the total radius to meters:
Total radius = 3790 km * 1000 = 3,790,000 meters

3. Substitute the values into the formula:
g = (6.67 × 10^-11 N m^2/kg^2) * (6.39 × 10^23 kg) / (3,790,000 meters)^2

4. Simplify the equation and calculate the result:
g = 2.967 m/s^2 (approximately)

So, the gravitational field strength of Mars at a position 400 km above its surface is approximately 2.967 m/s^2.