# What is the force of attraction between mars and the satellite?

Scientists want to place a 2900.0 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 1.6 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:

mmars = 6.4191 x 10^23 kg
rmars = 3.397 x 10^6 m
G = 6.67428 x 10^-11 N-m2/kg2

I know F = G * m * M /r^2

But what should I plug for r , m , M , G

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haruspex
Homework Helper
Gold Member
what should I plug for r , m , M , G
Start by making some suggestions that look reasonable to you.

haruspex
Homework Helper
Gold Member
what should I plug for r , m , M , G
What do you understand the various symbols to stand for in the equation?

What do you think ? I am worried about the radius because of the 1.6 times

m= 6.4191 x 10^23
G = is either 9.8 or 6.67428 x 10^-11
M= 2900
r=(1.6* 3.397 x 10^6) or 3.397 x 10^6

haruspex
Homework Helper
Gold Member
What do you think ? I am worried about the radius because of the 1.6 times

m= 6.4191 x 10^23
G = is either 9.8 or 6.67428 x 10^-11
M= 2900
r=(1.6* 3.397 x 10^6) or 3.397 x 10^6
There's not much point in knowing an equation if you don't know what the variables in it represent. In this one, G is the gravitational constant, not 'g', the acceleration due to gravity at earth's surface. The dimensions of the two are quite different. If you track the units through your working (and it is a very good idea to do that) you'll see that using 9.81 m/s2 would give the wrong dimensions for force in the answer.
r is the distance between the masses. In principle, this should refer only to point masses. Gravitational attraction between two bodies of arbitrary shape can get pretty nasty. But luckily, the equation also works when the bodies are spherical and uniformly dense*. In this case, you take r to be the distance between the mass centres.
[*In fact, they don't need to be uniformly dense. It is enough that, in each, the density is only a function of distance from the centre.]