An earth satelites S has orbit radius which $4$ times that of communication satellite C. The period of revolution of S will be.

Question

Verified by Toppr

Answer 1

Larger the distance of planet from the sun, larger will be its period of revolution around the sun.
Kepler's third low planetary motion the square of the period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.
$∴ T^{2} \propto R^{3}$
$\frac{T _{s}}{T _{c}} = (\frac{R _{s}}{R _{c}})^{3 / 2}$
given $R_{s} = 4 R_{c}$
$\frac{T _{s}}{T _{c}} (\frac{4 R _{c}}{R _{c}})^{3 / 2} = 8$
for $T_{c} = 1$ day
$T_{s} = 8$ days.

Answer 2

Larger the distance of planet from the sun, larger will be its period of revolution around the sun.
Kepler's third low planetary motion the square of the period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.

∴ T^{2} \propto R^{3}

\frac{T _{s}}{T _{c}} = (\frac{R _{s}}{R _{c}})^{3 / 2}

given

R_{s} = 4 R_{c}

\frac{T _{s}}{T _{c}} (\frac{4 R _{c}}{R _{c}})^{3 / 2} = 8

for

T_{c} = 1

day

T_{s} = 8

days.

An earth satelites S has orbit radius which $4$ times that of communication satellite C. The period of revolution of S will be.

$32$ days

$18$ days

$8$ days

$9$ days

The time period of a geostationary satellite at a height $36000 k m$ , is $24 h$ . A spy satellite orbits very close to earth surface ( $R = 6400 k m$ ). What will be its time period ?

A geostationary satellite is orbiting the earth at a height $6 R$ above the surface of earth , where $R$ is the radius of the earth. The time period of another satellite at a height of $2.5$ R from the surface of earth in hours is

A satellite is rotating around a planet in the orbi of radius r with time period T. If gravitational force changes according to $r^{5 / 2}$ the $T^{2}$ will be

A geostationary satellite is revolving at a height 6R above the earth's surface, where R is the radius of earth. The period of revolution of satellite orbiting at a height 2.5 R above the earth's surface will be:-

The time period of an orbiting satellite is (assuming spherical earth)

An earth satelites S has orbit radius which 4 times that of communication satellite C. The period of revolution of S will be.

32 days

18 days

8 days

9 days

The time period of a geostationary satellite at a height 36000 km, is 24 h. A spy satellite orbits very close to earth surface (R=6400 km). What will be its time period ?

A geostationary satellite is orbiting the earth at a height 6R above the surface of earth , where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of earth in hours is

A satellite is rotating around a planet in the orbi of radius r with time period T. If gravitational force changes according to r5/2 the T2 will be

A geostationary satellite is revolving at a height 6R above the earth's surface, where R is the radius of earth. The period of revolution of satellite orbiting at a height 2.5 R above the earth's surface will be:-

The time period of an orbiting satellite is (assuming spherical earth)

An earth satelites S has orbit radius which $4$ times that of communication satellite C. The period of revolution of S will be.

$32$ days

$18$ days

$8$ days

$9$ days

The time period of a geostationary satellite at a height $36000 k m$ , is $24 h$ . A spy satellite orbits very close to earth surface ( $R = 6400 k m$ ). What will be its time period ?

A geostationary satellite is orbiting the earth at a height $6 R$ above the surface of earth , where $R$ is the radius of the earth. The time period of another satellite at a height of $2.5$ R from the surface of earth in hours is

A satellite is rotating around a planet in the orbi of radius r with time period T. If gravitational force changes according to $r^{5 / 2}$ the $T^{2}$ will be