Question
5) What is the gravitational potential energy (relative to infinite) of a 5.00times 10^3kg satellite that is in orbit with a radius of 9.90times 10^6 m around the Earth? (-2.0times 10^11J) 6) How much work is done against gravity in lifting the satellite in probler #5 to its orbital height? (1.11times 10^11J)
Solution
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Macy
Professional · Tutor for 6 years
Answer
Here's how to solve these problems:**5) Gravitational Potential Energy*** **Formula:** The gravitational potential energy (U) of an object relative to infinity is given by: U = -GMm/r Where: * G is the gravitational constant (6.674 x 10⁻¹¹ Nm²/kg²) * M is the mass of the Earth (5.972 x 10²⁴ kg) * m is the mass of the satellite (5.00 x 10³ kg) * r is the distance between the center of the Earth and the satellite (9.90 x 10⁶ m)* **Calculation:** U = -(6.674 x 10⁻¹¹ Nm²/kg²) * (5.972 x 10²⁴ kg) * (5.00 x 10³ kg) / (9.90 x 10⁶ m) U ≈ -2.02 x 10¹¹ J* **Answer:** The gravitational potential energy is approximately -2.0 x 10¹¹ J. The negative sign indicates that the satellite is bound to the Earth.**6) Work Done Against Gravity*** **Concept:** The work done against gravity to lift the satellite to its orbital height is equal to the *change* in gravitational potential energy. Since the initial potential energy (at the surface of the Earth) is more negative than the final potential energy (in orbit), the work done is positive.* **Formula:** Work (W) = ΔU = U_final - U_initial* **U_initial:** We need to calculate the gravitational potential energy at the Earth's surface. The radius of the Earth is approximately 6.371 x 10⁶ m. U_initial = -(6.674 x 10⁻¹¹ Nm²/kg²) * (5.972 x 10²⁴ kg) * (5.00 x 10³ kg) / (6.371 x 10⁶ m) U_initial ≈ -3.13 x 10¹¹ J* **Calculation:** W = (-2.02 x 10¹¹ J) - (-3.13 x 10¹¹ J) W ≈ 1.11 x 10¹¹ J* **Answer:** The work done against gravity is approximately 1.11 x 10¹¹ J. This is the energy required to lift the satellite to that height.