Question

What is the gravitational potential energy of Mars due to the Sun? Assume M _Mars =6.42×10 ^23 kg, M_Sun =2.0×10 ^30kg, and the orbital distance is r=2.28×10 ^8 km. Express your answer with the appropriate units.

Answer 1

To calculate the gravitational potential energy of Mars due to the Sun, we can use the formula (G * MMars * MSun) / r, where G is the gravitational constant, MMars is the mass of Mars, MSun is the mass of the Sun, and r is the orbital distance. Plugging in the values and simplifying the calculation will give us the gravitational potential energy.

Step-by-step explanation:

Gravitational potential energy is the energy possessed by an object due to its position in a gravitational field. To calculate the gravitational potential energy of Mars due to the Sun, we can use the formula:

Potential energy = (G * MMars * MSun) / r

where G is the gravitational constant (6.67 x 10-11 N m2 / kg2), MMars is the mass of Mars (6.42 x 1023 kg), MSun is the mass of the Sun (2.0 x 1030 kg), and r is the orbital distance (2.28 x 108 km).

Plugging in the values, we have:

Potential energy = (6.67 x 10-11 N m2 / kg2) * (6.42 x 1023 kg * 2.0 x 1030 kg) / (2.28 x 108 km)

Converting km to meters, we get:

Potential energy = (6.67 x 10-11 N m2 / kg2) * (6.42 x 1023 kg * 2.0 x 1030 kg) / (2.28 x 108 km * 103 m/km)

Simplifying the calculation gives us the gravitational potential energy of Mars due to the Sun.