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Gravitational Field Strength in Circular Orbit

Created by @nathanedwards

at November 4th 2023, 7:11:34 pm.

AP_Physics_1-Questions

#gravitational-force

#gravitational-field-strength

#circular-orbit

AP Physics 1 Exam Question - Gravitational Force and Fields

A spacecraft of mass $m$ is in a circular orbit around a planet of mass $M$ and radius $r$ . The spacecraft is moving at a constant speed, which allows it to maintain its circular orbit. The gravitational force between the spacecraft and the planet can be given by the equation:

F = \frac{G \cdot M \cdot m}{r^2}

Where:

$F$ is the gravitational force between the spacecraft and the planet.
$G$ is the gravitational constant ( $6.674 \times 10^{-11}$ N m $^2$ /kg $^2$ ).

a) Derive an expression for the gravitational field strength at a point on the orbit of the spacecraft in terms of $M$ , $r$ , and $G$ .

b) Calculate the gravitational field strength for a spacecraft in a circular orbit around Earth at an altitude of 300 km, assuming a mass of $2 \times 10^4$ kg for the spacecraft and a mass of $5.97 \times 10^{24}$ kg for Earth.

Answer:

a) Deriving the expression for gravitational field strength:

The gravitational field strength at a point is defined as the gravitational force experienced by a unit mass placed at that point.

The formula for gravitational field strength, $g$ , is given by:

g = \frac{F}{m}

We can substitute the expression for gravitational force, $F$ , into the equation above:

g = \frac{\frac{G \cdot M \cdot m}{r^2}}{m}

Canceling out the mass $m$ on both numerator and denominator, we get:

g = \frac{G \cdot M}{r^2}

Hence, the expression for gravitational field strength at a point on the orbit of the spacecraft is:

g = \frac{G \cdot M}{r^2}

b) Calculating the gravitational field strength:

Given:

Mass of spacecraft, $m = 2 \times 10^4$ kg
Mass of Earth, $M = 5.97 \times 10^{24}$ kg
Radius, $r = 6,371$ km (radius of Earth + altitude)

First, convert the radius to meters:

r = 6,371 \times 1000 = 6.371 \times 10^6 \, \text{m}

Now, substitute the values into the expression for gravitational field strength:

g = \frac{G \cdot M}{r^2}

g = \frac{(6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (5.97 \times 10^{24} \, \text{kg})}{{(6.371 \times 10^6 \, \text{m})}^2}

g = \frac{(6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (5.97 \times 10^{24} \, \text{kg})}{4.053 \times 10^{13} \, \text{m}^2}

g = \frac{3.972 \times 10^{14}}{4.053 \times 10^{13}} \, \text{N/kg}

g \approx 9.81 \, \text{N/kg}

Therefore, the gravitational field strength for a spacecraft in a circular orbit around Earth at an altitude of 300 km is approximately $9.81 \, \text{N/kg}$ .