Question

  

A satellite has a mass of 3410 kg and is in a circular orbit 5.51 x 10⁶ m above the surface of a planet. The period of the orbit is 7.06 hours. The radius of the planet is 4.04 x 10⁶ m. What is the true weight of the satellite when it is at rest on the planet's surface?

Answer 1

The solution is as follows :

From the given problem, following data are obtained:

Ø The mass of the satellite (m) = 3410 kg.

Ø The distance of the satellite from the surface of the planet, (d) =5.51 × 10⁶ m.

Ø The radius of the planet, (R) = 4.04 × 10⁶ m.

Ø The period of the orbit, (T) = 7.06 hrs = 7.06 × 3600 = 25416 s = 2.5416 × 10⁴ s.

Let, ‘r’ be the orbital radius of the satellite.

Now, the orbital radius of the satellite is calculated as follows :

r = R + d

= 4.04 × 10⁶ + 5.51 × 10⁶

= 9.55 × 10⁶ m

The time period of the motion of a satellite around a planet depends only on the mass M, of the planet and its distance r, from the center of the planet buy the following equation:

T = 2π√(r³/ GM)

Here, G is the universal gravitational constant and its value is 6.67 × 10^-11 m³/kg-s²

Using the above equation of time period (T), the mass of the planet (M) can be obtained as follows:

Mass of the planet, (M)

= (2π / T)² × (r³/ G)

= (2 × 3.14 / 2.5416 × 10⁴)² × {(9.55 × 10⁶)³ / 6.67 × 10^-11}

= ( 2.47 × 10^-4)² × (130.58 × 10²⁹)

= 796.66 × 10²¹

= 0.796 × 10²⁴ kg.

Now, the acceleration due to gravity (g) at or outside the surface of the planet is given by the following equation:

g = GM / R²

where, M = mass of the planet;

and, R = the distance of the point from the center of the planet = radius of the planet

[Because at the surface of the planet, the radius of the planet (R) = the distance of the point from the center of the planet.]

So, the acceleration due to gravity ‘g’ is

= (6.67 × 10^-11 × 0.796 × 10²⁴) / (4.04 × 10⁶)²

= 3.25 m/s²

The of weight (W) of the satellite is given by the equation: W = mg;

where, m = mass of the satellite and g = acceleration due to gravity.

Hence, the weight of the satellite on the surface of the planet is given by,

W = 3410 kg × 3.25 m/s² = 11.1 kN = 1.11 × 10⁴ N

Answer: The required true weight of the satellite when it is at rest on the surface of the planet is 1.11 × 10⁴ N