Question

The radius of the earth's very nearly circular orbit around the sun is 1.5x10^11m.

Find the magnitude of the earth's velocity as it travels around the sun. Assume a year of 365 days.

Find the magnitude of the earth's angular velocity as it travels around the sun. Assume a year of 365 days.

Find the magnitude of the earth's centripetal acceleration as it travels around the sun. Assume a year of 365 days.

Answer 1

Concepts and reason

The concepts used in this problem are orbital velocity, relation between angular velocity and linear velocity and the centripetal acceleration.

The Earth’s velocity can be calculated using the expression of orbital velocity. The angular velocity of Earth can be calculated suing the relation between angular velocity and linear velocity. The centripetal acceleration can be calculated using the expression of centripetal acceleration.

Fundamentals

Orbital velocity:

When an astronomical object moves in an orbit around another astronomical object, then its velocity is given by the below expression.

$v = \sqrt {\frac{{GM}}{r}}$

Here, $G$is the gravitational constant, $M$is the mass of that object around which the other object is moving and $r$is the radius of orbit.

Centripetal acceleration:

The linear acceleration is the rate of change of linear velocity but the rate of change of tangential velocity is the centripetal acceleration. The centripetal makes an object to move in a curved path.

The expression for centripetal acceleration is:

$a = \frac{{{v^2}}}{r}$

Here, $a$is the centripetal acceleration, $v$is the velocity and $r$is the radial distance.

Relation between angular speed and linear speed:

The angular speed is:

$\omega = \frac{v}{r}$

Here, $r$is the radial distance.

(a)

The expression for velocity is:

$v = \sqrt {\frac{{GM}}{r}}$

The mass of Sun is$1.99 \times {10^{30}}{\rm{ kg}}$and the value of gravitational constant is$6.67 \times {10^{ - 11}}{\rm{ N}}{{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}$.

Substitute$1.99 \times {10^{30}}{\rm{ kg}}$for$M$, $6.67 \times {10^{ - 11}}{\rm{ N}}{{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}$for $G$and $1.5 \times {10^{11}}{\rm{ m}}$for$r$in the above expression.

$\begin{array}{c}\\v = \sqrt {\frac{{\left( {6.67 \times {{10}^{ - 11}}{\rm{ N}}{{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right)\left( {1.99 \times {{10}^{30}}{\rm{ kg}}} \right)}}{{1.5 \times {{10}^{11}}{\rm{ m}}}}} \\\\ = \sqrt {884.88 \times {{10}^6}{\rm{ }}{{\rm{m}}^2}/{{\rm{s}}^2}} \\\\ = 2.97 \times {10^4}{\rm{ m/s}}\\\end{array}$

(b)

The expression for angular velocity is:

$\omega = \frac{v}{r}$

Substitute$2.97 \times {10^4}{\rm{ m/s}}$for$v$and $1.5 \times {10^{11}}{\rm{ m}}$for$r$in the above expression.

$\begin{array}{c}\\\omega = \frac{{2.97 \times {{10}^4}{\rm{ m/s}}}}{{1.5 \times {{10}^{11}}{\rm{ m}}}}\\\\ = 1.98 \times {10^{ - 7}}{\rm{ rad/s}}\\\end{array}$

(c)

The expression of centripetal acceleration is:

$a = \frac{{{v^2}}}{r}$

Substitute$2.97 \times {10^4}{\rm{ m/s}}$for$v$and $1.5 \times {10^{11}}{\rm{ m}}$for$r$in the above expression.

$\begin{array}{c}\\a = \frac{{{{\left( {2.97 \times {{10}^4}{\rm{ m/s}}} \right)}^2}}}{{1.5 \times {{10}^{11}}{\rm{ m}}}}\\\\ = \frac{{8.84 \times {{10}^8}{\rm{ }}{{\rm{m}}^2}/{{\rm{s}}^2}}}{{1.5 \times {{10}^{11}}{\rm{ m}}}}\\\\ = 5.89 \times {10^{ - 3}}{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Ans: Part a

The magnitude of Earth’s velocity is${\bf{2}}{\bf{.97 \times 1}}{{\bf{0}}^{\bf{4}}}{\bf{ m/s}}$.

Part b

The magnitude of angular velocity of Earth is${\bf{1}}{\bf{.98 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ rad/s}}$.

Part c

The magnitude of centripetal acceleration of Earth is${\bf{5}}{\bf{.89 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{ m/}}{{\bf{s}}^{\bf{2}}}$.