Post

Calculating Gravitational Force and Orbital Speed

Created by @nathanedwards
 at November 2nd 2023, 4:39:56 pm.
AP_Physics_1-Questions
#gravitational-force
#gravitational-field-strength
#orbital-speed

Question:

A small satellite with a mass of 100 kg is orbiting around a planet with a mass of 5×10^24 kg. The satellite is at a distance of 100,000 km from the center of the planet.

a) Calculate the magnitude of the gravitational force experienced by the satellite.

b) Determine the gravitational field strength at the location of the satellite.

c) If the satellite is at rest, calculate the orbital speed required for it to remain in orbit at this distance.

Assume the gravitational constant, G, is 6.673 × 10^-11 N m^2/kg^2.

Answer:

a) To calculate the magnitude of the gravitational force experienced by the satellite, we can use the formula for gravitational force:

F=G⋅m1⋅m2r2 F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}}

where

F is the gravitational force, G is the gravitational constant (6.673×10−116.673 × 10^{-11} N m²/kg²), m1m_1 is the mass of the satellite (100 kg), m2m_2 is the mass of the planet (5×10245 × 10^{24} kg), and r is the distance between the satellite and the center of the planet (100,000 km = 10510^5 m).

Substituting the given values into the formula, we have:

F=(6.673×10−11 N m2/kg2)⋅(100 kg)⋅(5×1024 kg)(105 m)2 F = \frac{{(6.673 × 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (100 \, \text{kg}) \cdot (5 × 10^{24} \, \text{kg})}}{{(10^5 \, \text{m})^2}}

Simplifying the calculation:

F=3325×10131010=3325×103=3.325×106 N F = \frac{{3325 \times 10^{13}}}{{10^{10}}} = 3325 \times 10^3 = 3.325 × 10^6 \, \text{N}

Therefore, the magnitude of the gravitational force experienced by the satellite is 3.325×106 N3.325 × 10^6 \, \text{N}.

b) To determine the gravitational field strength at the location of the satellite, we can use the formula:

g=Fm g = \frac{F}{m}

where g is the gravitational field strength, F is the gravitational force on the satellite (found in part a, 3.325×106 N3.325 × 10^6 \, \text{N}), and m is the mass of the satellite (100 kg).

Plugging in the values:

g=3.325×106 N100 kg=3.325×104 N/kg g = \frac{3.325 × 10^6 \, \text{N}}{100 \, \text{kg}} = 3.325 × 10^4 \, \text{N/kg}

Therefore, the gravitational field strength at the location of the satellite is 3.325×104 N/kg3.325 × 10^4 \, \text{N/kg}.

c) To calculate the orbital speed required for the satellite to remain in orbit at this distance, we can use the formula for orbital speed:

v=G⋅Mr v = \sqrt{\frac{{G \cdot M}}{{r}}}

where v is the orbital speed, G is the gravitational constant (6.673×10−116.673 × 10^{-11} N m²/kg²), M is the mass of the planet (5×10245 × 10^{24} kg), and r is the distance between the satellite and the center of the planet (100,000 km = 10510^5 m).

Substituting the given values into the formula, we have:

v=(6.673×10−11 N m2/kg2)⋅(5×1024 kg)(105 m) v = \sqrt{\frac{{(6.673 × 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (5 × 10^{24} \, \text{kg})}}{{(10^5 \, \text{m})}}}

Simplifying the calculation:

v=3325×1013105=3325×108=3.325×1012 m/s v = \sqrt{\frac{{3325 \times 10^{13}}}{{10^5}}} = \sqrt{3325 \times 10^8} = \sqrt{3.325 × 10^{12}} \, \text{m/s}

Therefore, the orbital speed required for the satellite to remain in orbit at this distance is approximately 1.825×106 m/s1.825 \times 10^6 \, \text{m/s}.

Chat

Save To Bookmark