Home / CIE AS & A Level Physics 9702: Topic 13: Gravitational fields- Unit : 13.2 Gravitational force between point masses Study Notes

CIE AS & A Level Physics 9702: Topic 13: Gravitational fields- Unit : 13.2 Gravitational force between point masses Study Notes

Newton’s Law of Gravitation

  • The gravitational force between two bodies outside a uniform field e.g. between the Earth and the Sun, is defined by Newton’s Law of Gravitation
  •  Recall that the mass of a uniform sphere can be considered to be a point mass at its centre
  •  Newton’s Law of Gravitation states that:
                        The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square their separation
  •  In equation form, this can be written as:

$
\mathrm{F}_{\mathrm{G}}=\frac{G m_1 m_2}{r^2}
$

                     

The gravitational force between two masses outside a uniform field is defined by Newton’s Law of Gravitation

  •  Where:
    •  $\mathrm{F}_{\mathrm{G}}=$ gravitational force between two masses ( $\mathrm{N}$ )
    • $\mathrm{G}=$ Newton’s gravitational constant
    • $\mathrm{m}_1$ and $\mathrm{m}_2=$ two points masses $(\mathrm{kg})$
    •  $r=$ distance between the centre of the two masses ( $m$ )
  • Although planets are not point masses, their separation is much larger than their radius
    • Therefore, Newton’s law of gravitation applies to planets orbiting the Sun
  •  The $1 / r^2$ relation is called the ‘inverse square law’
  • This means that when a mass is twice as far away from another, its force due to gravity reduces by $(1 / 2)^2=1 / 4$

Worked example: Newton’s law of gravitation
A satellite with mass $6500 \mathrm{~kg}$ is orbiting the Earth at $2000 \mathrm{~km}$ above the Earth’s surface. The gravitational force between them is $37 \mathrm{kN}$.
Calculate the mass of the Earth.
Radius of the Earth $=6400 \mathrm{~km}$.

Answer/Explanation

STEP 1 NEWTON’S LAW OF GRAVITATION
$
F_G=\frac{G m_1 m_2}{r^2}
$
$m_1$ IS THE MASS OF THE SATELLITE $\Leftarrow \begin{aligned} & \text { THESE CAN BE } \\ & m_2 \text { IS THE MASS OF THE EARTH }\end{aligned}$ ANY WAY AROUND
STEP 2 REARRANGE FOR $m_2$ (MASS OF EARTH)
$
\frac{r^2 F_G}{G m_1}=m_2

STEP 3  CALCULATE THE DISTANCE $r$
r IS THE DISTANCE BETWEEN THE CENTRE OF THE EARTH AND SATELLITE
$r=$ DISTANCE OF SATELLITE ABOVE THE SURFACE + RADIUS OF THE EARTH

     

$r=2000+6400=8400 \mathrm{~km}=8400 \times 10^3 \mathrm{~m}$
STEP 4 NEWTON’S GRAVITATIONAL CONSTANT

SUBSTITUTE IN VALUES

$\frac{\left(8400 \times 10^3\right)^2 \times 37 \times 10^3}{6.67 \times 10^{-11} \times 6500}=6.0 \times 10^{24} \mathrm{~kg} \quad(2$ s.f.) 

$37 \mathrm{kN}$

Exam Tip
A common mistake in exams is to forget to add together the distance from the surface of the planet and its radius to obtain the value of $r$. The distance $r$ is measured from the centre of the mass, which is from the centre of the planet.

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