Answer:
a. To find the force of gravity acting on Phobos, we can use Newton's Law of Gravitation, which states that the force of gravity between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The equation is F = G(m1m2/d^2), where F is the force of gravity, G is the gravitational constant (6.674 × 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the two objects, and d is the distance between their centers. Plugging in the values for Mars, Phobos, and their distance apart, we get:
F = (6.674 × 10^-11 Nm^2/kg^2)(6.4 × 10^23 kg)(1.07 × 10^16 kg)/(9.4 × 10^6 m)^2
F = 9.91 × 10^15 N
b. To find the force of gravity acting on the scientific satellite, we can use the equation for the gravitational force between two objects again, but this time we need to take into account the fact that the satellite is 200 km above the surface of Mars. The distance between the center of Mars and the satellite is therefore the sum of the radius of Mars (3396 km) and the altitude of the satellite (200 km), or 3596 km. Plugging in the values, we get:
F = (6.674 × 10^-11 Nm^2/kg^2)(6.4 × 10^23 kg)(1300 kg)/(3.596 × 10^6 m)^2
F = 398 N
c. The gravitational potential energy of the satellite in a circular orbit around Mars is given by the equation U = -G(m1m2)/r, where U is the gravitational potential energy, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers. Since the distance between the satellite and Mars remains constant in a circular orbit, the gravitational potential energy of the satellite also remains constant.
d. The acceleration of the satellite in a circular orbit around Mars is given by the equation a = v^2/r, where a is the acceleration, v is the velocity of the satellite, and r is the distance between the satellite and Mars. Since the distance between the satellite and Mars remains constant in a circular orbit, the acceleration of the satellite also remains constant.