Worked examples · Sheet 2

Gravitational acceleration on other planets, at altitude and at depth

Theory used: Gravitational acceleration at the surface Variation of g with altitude Variation of g with depth Field of a homogeneous sphere Weightlessness in orbit

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Mars has mass $M = 6.42 \cdot 1 0^{23} kg$ and radius $R = 3.39 \cdot 1 0^{6} m$ . Compute the gravitational acceleration at its surface and the weight on Mars of an astronaut of mass $m = 70 kg$ , comparing it with the weight on the Earth ( $g_{0} = 9.81 m/ s^{2}$ ). Neglect the rotation of the planet.

g_{M} = \frac{GM}{R ^{2}}, P_{M} = m g_{M}

g_{M} = \frac{6.67 \cdot 1 0 ^{- 11} \cdot 6.42 \cdot 1 0 ^{23}}{( 3.39 \cdot 1 0 ^{6} ) ^{2}} m/ s^{2} = \frac{4.28 \cdot 1 0 ^{13}}{1.149 \cdot 1 0 ^{13}} m/ s^{2} = 3.73 m/ s^{2}

P_{M} = 70 \cdot 3.73 N = 261 N, P_{T} = 70 \cdot 9.81 N = 687 N, \frac{P _{M}}{P _{T}} = \frac{g _{M}}{g _{0}} = 0.380

The planet is considered a sphere whose density depends only on the distance from the center, so at the surface it attracts like a point mass placed at the center at distance $R$ . The gravitational acceleration at the surface is therefore $\frac{GM}{R ^{2}}$ . Neglecting the rotation, the weight coincides with the gravitational force $m g_{M}$ .

The mass of the astronaut is the same on the two planets, while the weight changes in the ratio of the gravitational accelerations: on Mars the astronaut weighs $38%$ of what they weigh on the Earth. As an order-of-magnitude check, Mars has about one tenth of the Earth’s mass and about half its radius; since $g$ is proportional to $\frac{M}{R ^{2}}$ , we get about $\frac{0.1}{0. 5 ^{2}} = 0.4$ times the Earth’s value, in agreement with the calculation.

Exercise 2

Determine the altitude $h$ at which the Earth’s gravitational acceleration equals one quarter of its value at the surface and the altitude at which it has decreased by $1%$ . For the second altitude compare the exact value with the one given by the approximation for small altitudes. Use $R_{T} = 6.37 \cdot 1 0^{6} m$ .

g (h) = g_{0} \frac{R _{T}^{2}}{( R _{T} + h ) ^{2}} = k g_{0} ⟺ \frac{R _{T}}{R _{T} + h} = k ⟺ h = R_{T} (\frac{1}{k} - 1)

k = \frac{1}{4} : h = R_{T} (2 - 1) = R_{T} = 6.37 \cdot 1 0^{6} m

k = 0.99 : h = 6.37 \cdot 1 0^{6} (\frac{1}{0.99} - 1) m = 6.37 \cdot 1 0^{6} \cdot 5.04 \cdot 1 0^{- 3} m = 3.21 \cdot 1 0^{4} m

g \approx g_{0} (1 - \frac{2 h}{R _{T}}) = 0.99 g_{0} ⟺ h = 0.005 R_{T} = 3.19 \cdot 1 0^{4} m

\frac{g}{g _{0}}

as a function of

\frac{h}{R _{T}}

: at the altitude

h = R_{T}

the ratio is

\frac{1}{4}

; the dashed line is the approximation

1 - \frac{2 h}{R _{T}}

, valid only near

h = 0

By the variation of g with altitude the ratio $\frac{g}{g _{0}}$ is $\frac{R _{T}^{2}}{( R _{T} + h ) ^{2}}$ . We set it equal to $k$ and take the square root; both sides are positive, so the resulting equation is equivalent and is solved for $h$ . With $k = \frac{1}{4}$ the distance from the center must double, therefore the altitude is equal to the Earth’s radius.

For a decrease of $1%$ the exact altitude is $32.1 km$ . The first-order approximation gives $31.9 km$ , with a difference of less than $1%$ , because the altitude is small compared with $R_{T}$ . The graph shows that at altitudes comparable with $R_{T}$ the line moves away from the exact curve and the approximation does not apply.

Exercise 3

The International Space Station orbits at an altitude $h = 400 km$ . Compute the gravitational acceleration at that altitude, in absolute value and as a ratio to the value $g_{0} = 9.81 m/ s^{2}$ at the surface, with $R_{T} = 6.37 \cdot 1 0^{6} m$ , and explain why the astronauts on board float.

g (h) = g_{0} (\frac{R _{T}}{R _{T} + h})^{2}

\frac{g ( h )}{g _{0}} = (\frac{6.37 \cdot 1 0 ^{6}}{6.77 \cdot 1 0 ^{6}})^{2} = 0.94 1^{2} = 0.885

g (h) = 0.885 \cdot 9.81 m/ s^{2} = 8.69 m/ s^{2}

The altitude is added to the Earth’s radius to obtain the distance from the center, $6.77 \cdot 1 0^{6} m$ , and the variation of g with altitude gives a gravitational acceleration equal to $88.5%$ of the value at the ground. The gravitational force on the astronauts is therefore almost the same as the one they feel on the Earth.

The astronauts float because the station and everything it contains move with the same acceleration $g$ , directed toward the center of the Earth: they are in free fall together. The floor of the station does not have to support the astronauts, so it exerts no normal force on them, and the constraint reaction is what is perceived as weight. Therefore the apparent weight is zero even though the gravitational force is present.

Exercise 4

In the model of a spherical Earth of uniform density, determine the depth $d$ at which the gravitational acceleration equals that at the altitude $h = 100 km$ above the surface. Use $R_{T} = 6.37 \cdot 1 0^{6} m$ and compare the result with $2 h$ .

g_{depth} (d) = g_{0} (1 - \frac{d}{R _{T}}), g_{alt} (h) = g_{0} \frac{R _{T}^{2}}{( R _{T} + h ) ^{2}}

1 - \frac{d}{R _{T}} = \frac{R _{T}^{2}}{( R _{T} + h ) ^{2}} ⟺ d = R_{T} (1 - \frac{R _{T}^{2}}{( R _{T} + h ) ^{2}})

\frac{R _{T}^{2}}{( R _{T} + h ) ^{2}} = (\frac{6.37}{6.47})^{2} = 0.9693

d = 6.37 \cdot 1 0^{6} \cdot (1 - 0.9693) m = 1.95 \cdot 1 0^{5} m

Inside a homogeneous sphere the field grows linearly with the distance from the center, so at depth $d$ , that is, at distance $R_{T} - d$ from the center, it equals $g_{0} (1 - \frac{d}{R _{T}})$ ; this is the variation of g with depth. Outside, instead, the $\frac{1}{r ^{2}}$ law holds. Equating the two expressions, $g_{0}$ cancels and we obtain $d$ .

The required depth is about $195 km$ , almost twice the altitude. The reason can be seen from the first-order approximations: at altitude $g \approx g_{0} (1 - \frac{2 h}{R _{T}})$ , at depth $g = g_{0} (1 - \frac{d}{R _{T}})$ , so the two accelerations coincide for $d \approx 2 h = 200 km$ . The small difference from $200 km$ is due to the terms neglected in the approximation.