Physics, asked by kanadsa2226, 3 months ago

calculate the radial velocity of planet X whose mass is twice of earth with gravitational strength 3/4 of Earth. it is revolving around a star of 5 solar mass with a luminosity of 4 ergs/s

Answers

Answered by Braɪnlyємρєяσя
3

Explanation:

The surface gravity, g, of an astronomical object is the gravitational acceleration experienced at its surface at the equator, including the effects of rotation. The surface gravity may be thought of as the acceleration due to gravity experienced by a hypothetical test particle which is very close to the object's surface and which, in order not to disturb the system, has negligible mass.

Surface gravity is measured in units of acceleration, which, in the SI system, are meters per second squared. It may also be expressed as a multiple of the Earth's standard surface gravity, g = 9.80665 m/s².[1] In astrophysics, the surface gravity may be expressed as log g, which is obtained by first expressing the gravity in cgs units, where the unit of acceleration is centimeters per second squared, and then taking the base-10 logarithm

Answered by HorridAshu
1

\huge\bold{\mathtt{\red{A{\pink{N{\green{S{\blue{W{\purple{E{\orange{R}}}}}}}}}}}}}

\small\mathbf\red{{Explaination\::-}}

The surface gravity, g, of an astronomical object is the gravitational acceleration experienced at its surface at the equator, including the effects of rotation. The surface gravity may be thought of as the acceleration due to gravity experienced by a hypothetical test particle which is very close to the object's surface and which, in order not to disturb the system, has negligible mass.

Surface gravity is measured in units of acceleration, which, in the SI system, are meters per second squared. It may also be expressed as a multiple of the Earth's standard surface gravity, g = 9.80665 m/s².[1] In astrophysics, the surface gravity may be expressed as log g, which is obtained by first expressing the gravity in cgs units, where the unit of acceleration is centimeters per second squared, and then taking the base-10 logarithm.

\huge\mathcal{\fcolorbox{lime}{black}{\pink{Hope it's help u}}}

Similar questions