Question

A spaceship orbits around a planet at a height of $20km$ from its surface. Assuming that only the gravitational field of the planet acts on the spaceship. What will be the number of complete revolutions made by the spaceship $24hours$ around the planet? [Given the mass of the plane $ = 8 \times {10^{22}}kg$ , the radius of the planet $ = 2 \times {10^6}m$]
Gravitational constant $G = 6.67 \times {10^{ - 11}}N{m^2}/k{g^2}$]
$\left( a \right)$$9$
$\left( b \right)$$11$
$\left( c \right)$$13$
$\left( d \right)$$17$

Answer 1

Hint Here in this question the height is given and we will assume in this question that there is only the gravitational pull acting on it. And we have to find the number of complete revolutions made by the ship during the given time period.
Formula:
Gravitational force,
$F = \dfrac{{Gmm}}{{{r^2}}}$
And
Centrifugal force,
$F = \dfrac{{m{v^2}}}{r}$
Where,
$F{{ and G}}$ , will be the centrifugal force and gravitational constant respectively.
$m$ , will be the mass
$V$ , will be the velocity and $r$ will be the distance between them.

Complete Step By Step Solution So in this question, the values which are given are the mass of the planet, their radius, and the height from which it’s moving in the orbit.
Hence from the question, it is clear that the two forces will act on the body one is the gravitational force that is inside the other one is a centrifugal force that pulls the body outside.
So there will be a balanced between these two forces developed and
Mathematically we can write this as,
$ \Rightarrow \dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}$
So from here, we will calculate the value of $V$
And it will be,
$ \Rightarrow V = \sqrt {\dfrac{{GM}}{r}} $
Now on substituting the values, we will get
$ \Rightarrow \sqrt {\dfrac{{6.67 \times {{10}^{ - 11}} \times 8 \times {{10}^{22}}}}{{2.02 \times {{10}^6}}}} $
So now we will solve the above equation and we will get the value for the $V$
We get,
$ \Rightarrow V = 1.625 \times {10^3}m/s$
So at the above velocity, the body is revolving around the planet.
Now we have to find the revolutions, so for this
$ \Rightarrow nT$
Here,
Time period, $T$
$ \Rightarrow T = \dfrac{{2\pi r}}{V}$
By using the revolution formula we get,
$ \Rightarrow nT = 24 \times 60 \times 60$
$ \Rightarrow n\left( {\dfrac{{2\pi r\pi }}{V}} \right) = 24 \times 3600$
On substituting the values we get
$ \Rightarrow n \times \dfrac{{2 \times 3.14 \times 2.02 \times {{10}^6}}}{{1.625 \times {{10}^3}}} = 24 \times 3600$
Further solving the above equation, we get
$ \Rightarrow n = 11$

Therefore the total of $11$ complete revolutions will be done. Hence, the option $b$ will be the right choice.

Note Centrifugal force is not a force like other forces such as gravitational or electric. It’s only a mathematical concept used to simplify problems of motion in non-inertial reference frames. This force is discovered to save Newton’s laws of motion in the original form. Newton’s laws hold only if the observer is in an inertial reference frame but fails when the observer is in a non-inertial reference frame.