NESA Physics Motion in Gravitational Fields

This describes Kepler's First Law, also known as the Law of Ellipses, which states that planets move in elliptical orbits with the Sun at one focus.

Kepler's laws are not derived from combining Newton's first law with uniform circular motion; rather, the third law relates orbital period to distance.

This describes Kepler's Second Law, or the Law of Equal Areas, which states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

A circular orbit requires a negative total energy and a launch parallel to the surface at a specific orbital velocity, which is less than the escape velocity.

An elliptical orbit requires the total energy of the system to be negative, meaning the launch velocity must be strictly less than the escape velocity.

While the trajectory is indeed a parabola, an object launched at or above escape velocity will overcome the planet's gravity and never return to land.

This assumes the orbital radius is inversely proportional to the square of the satellite's mass. However, orbital speed and radius are completely independent of the satellite's mass.

This incorrectly assumes the orbital radius is directly proportional to the satellite's mass. The mass of the orbiting object cancels out when equating gravitational and centripetal forces.

This incorrectly assumes the orbital radius is proportional to the square of the satellite's mass. A satellite's mass has no effect on its orbital radius for a given constant speed.

This incorrect value results from failing to square the period $T$ in Kepler's Third Law, calculating $\sqrt[3]{\frac{GMT}{4\pi^2}}$. Additionally, this radius is smaller than Earth's radius ($6.37 \times 10^6$ m), which is physically impossible for an orbit.

This answer is the result of a mathematical error where the square root is taken instead of the cube root when solving for $r$ (calculating $\sqrt{\frac{GMT^2}{4\pi^2}}$).

This value is far too large for an Earth orbit with such a short period (it is larger than the distance to the Sun) and results from a significant calculation error or use of incorrect constants.

This value is far too small for a star (even smaller than Earth). This result likely stems from failing to convert the orbital period from years to seconds before applying the formula.

This answer is incorrect. It does not match the result derived from Kepler's Third Law, possibly due to arithmetic errors in handling the powers of 10.

This value is significantly larger than the correct mass. It may result from multiplying by the gravitational constant $G$ instead of dividing, or other algebraic errors.

Incorrect. This value is obtained by multiplying Earth's escape velocity by $3/4$, incorrectly assuming escape velocity is proportional to $R/M$.

Incorrect. This value is obtained by multiplying Earth's escape velocity by $\sqrt{3/4}$, incorrectly assuming escape velocity is proportional to $\sqrt{R/M}$.

Incorrect. This value is obtained by multiplying Earth's escape velocity by $4/3$, incorrectly assuming escape velocity is proportional to $M/R$ instead of $\sqrt{M/R}$.

This value is incorrect because it is too close to the standard surface gravity ($9.81 \text{ m s}^{-2}$). This result implies that the altitude of $400 \text{ km}$ was ignored and the radius was taken simply as $R$.

This value is incorrect and results from using the altitude ($h = 400 \text{ km}$) as the distance $r$ in the denominator, rather than the distance from the Earth's center ($R + h$).

This value represents the magnitude of the gravitational potential ($V = -\frac{GM}{r}$) at that altitude, measured in $\text{J kg}^{-1}$, rather than the gravitational field strength.

Equating gravitational force to centripetal force yields $v = \sqrt{GM/r}$, not $v \propto r^2$. This option incorrectly suggests velocity increases rapidly as the orbit gets larger.

While gravitational force is inversely proportional to the square of $r$ ($F_g \propto 1/r^2$), orbital velocity depends on the square root of the inverse of $r$.

This would mean velocity increases as the radius increases ($v \propto \sqrt{r}$). In reality, satellites further away experience weaker gravity and travel slower.

This value is significantly smaller than the correct period. It likely results from a calculation error or incorrect unit conversions during the application of the orbital period formula.

This option is incorrect. It does not match the result derived from Kepler's Third Law using the standard mass of Earth and the gravitational constant.

This answer corresponds to calculating $T^2$ (in seconds squared) and dividing by 3600. This indicates a mistake where the square root was omitted in the final step of the calculation.

This value is incorrect and does not result from the proper application of Newton's Law of Universal Gravitation.

This result is obtained by multiplying the gravitational constant by the two masses but failing to divide by the distance ($F = G m_1 m_2$).

This calculation incorrectly divides by the distance $r$ instead of the distance squared $r^2$ ($F = G \frac{m_1 m_2}{r}$).