Question

If mass and radius of the Earth are $6.0\times {10}^{24}kg$ and $6.4\times {10}^{6}m$ respectively. Calculate the force exerted by the Earth on a body of mass 1 kg, acceleration of the body of mass 1 kg and acceleration of the Earth in order.
Use $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$

• A. $9.8N$
  $9.8m{s}^{-2}$
  $9.8m{s}^{-2}$
• B. $9.8N$
  $9.8m{s}^{-2}$
  $1m{s}^{-2}$
• C. $9.8N$
  $9.8m{s}^{-2}$
  $1.63\times {10}^{-24}m{s}^{-2}$
• D. $9.8N$
  $1.63\times {10}^{-24}m{s}^{-2}$
  $9.8m{s}^{-2}$

Answer 1

The correct option is C $9.8N$

$9.8m{s}^{-2}$
$1.63\times {10}^{-24}m{s}^{-2}$
From Newton's law of gravitation, the force of attraction between the bodies is given by:
$F=\frac{G{m}_1{m}_2}{r^2}$
Mass of Earth, $m_1=6.0\times {10}^{24}$ kg
Mass of the body, $m_2=1kg$
Distance between the two bodies here is equal to the radius of earth.
$r=6.4\times {10}^6$ m
Gravitational constant, $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$

Substituting values, we get
$F=\frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}\times 1}{{(6.4\times {10}^6)}^2}$
$F=9.8N$

Using newton's second law of motion, acceleration of the body is given by:
$a_2=\frac{F}{m_2}$
$a_2=\frac{9.8N}{1kg}=9.8m{s}^{-2}$

Similarly, acceleration of the Earth is given by:
$a_1=\frac{F}{m_1}$
$a_1=\frac{9.8N}{6.0\times {10}^{24}kg}$
$a_1=1.63\times {10}^{-24}$ $m{s}^{-2}$