Question:medium

If the orbital speed of a body revolving in a circular path near the surface of the earth is 8 kms$^{-1}$, then the orbital speed of a body revolving around the earth in a circular orbit at height of 19,200 km from the surface of earth is (Radius of the earth = 6400 km)

Show Hint

For problems comparing orbital speeds at different altitudes, the proportionality $v_o \propto 1/\sqrt{r}$ (where $r$ is the distance from the planet's center, not the altitude) is the most efficient tool. Set up a ratio to cancel out the constants $G$ and $M$.

Updated On: Mar 30, 2026

4 kms$^{-1}$
6 kms$^{-1}$
7.5 kms$^{-1}$
9 kms$^{-1}$

Show Solution

The Correct Option is A

Solution and Explanation

Was this answer helpful?

0

Top Questions on Gravitation

Questions Asked in TS EAMCET exam

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

TS EAMCET - 2025
Complex numbers

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

TS EAMCET - 2025
Matrices and Determinants

If \( p_1, p_2, p_3 \) are the altitudes and \( a=4, b=5, c=6 \) are the sides of a triangle ABC, then \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \)

TS EAMCET - 2025
Geometry

If \(\alpha, \beta, \gamma, \delta, \epsilon\) are the roots of the equation \(x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36 = 0\) and \(\alpha<\beta<\gamma<\delta<\epsilon\) then \( \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} = \)

TS EAMCET - 2025
Algebra