Force of gravity? Acceleration?

Question

Mars has a mass of 6.4 × 1023 kg. It has a moon named Phobos with a mass of 1.07 × 1016 kg.

a. If their centers are 9400 km apart, what is the force of gravity acting on Phobos?

b. If the radius of Mars is 3.37 × 106 m, what is the acceleration of gravity on the surface of Mars?

c. What is the acceleration of Phobos as it orbits Mars?

Answer 1

Newtons Gravitational law will help us solve this problem!

The gravitational force between two masses is a function of their product divided by the squared distance between their centers of mass. 

F = G * (M1*M2)/r^2

where r is the distance between the masses, M1 and M2 are the masses of the two objects, and G is newtons gravitational constant G = 6.674×10^−11 N⋅m^2/kg^2

 

Part a)

F = G*M_mars*M_phobos/r^2

*units! G and M are in standard kg, m, s, so the distance (given in kilometers) should also be in meters. 

F = (6.674×10^−11) (6.4*10^24)(1.07*10^16)/(9400000^2)

F = 5.17*10^15 N 

 

Part b)

Here we relate the previous gravitational law to Newton's second law F = ma to solve for the acceleration of an unknown mass on the surface of Mars. The radius of the planet is the distance between the two objects.

 

F = G * (M_mars*M_unknown)/r^2 = M_unknown*a

(note that what we are doing here is the sum of all forces on an unknown mass is equal to the product of the unknown mass and it's acceleration, ie its freefall acceleration if there were no other forces but gravity acting on it.

Simply divide by the unknown mass and you solve for the acceleration.

a = G * (M_mars)/r^2

a = (6.674×10^−11) (6.4*10^24)/((3.37*10^6)^2)

a = 3.76 m/s^2 (about a third of that on Earth-this is the same process used to obtain the classic value for g = 9.8)

 

Part c)

This is a simple restatement of F=ma with the answer from part a. We already solved the force of gravity on phobos due to mars, and the mass of phobos is given, therefore:

F_gravity = M_phobos*a_phobos

a_phobos = (5.17*10^15)/(1.07*10^16)

a_phobos = 0.483 m/s

(note that this is your circular acceleration, meaning it is directed radially inward and is equal to v^2/r, so we can even back out the orbital speed of phobos of about 2130 m/s)

Answer 2

a) Let mass of mars is represented here as M_m and mass of phobos is as M_{p ,} the distance between their centers is as R,

 

According to Newton's law of universal gravitation,

 

F = G M_m M_p/ R²

 

put all the values in it and calculate, here G = 6.674 x 10^-11 N. (m/kg)²

 

F = (6.674 x 10^-11 x 6.4 x 10²³ x 1.07 x 10¹⁶⁾ / (9400 x 10³)^{2   N}   ( change distance in meters )

 

 

b) Let acceleration of gravity on the surface of mars is 'g' and radius of mars is represented as 'r'

 

g = G M_m / r²

 

= (6.674 x 10^-11 x 6.4 x 10²³ ) / ( 3.37 x 10⁶ )² m/s²